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Course Script:

Elementary Algebra, Functions, and Statistics

Vol. 2: Applications 2nd Edition

by Prof. Dr. Dietrich Ohse

ProCredit Academy GmbH Hammelbacher Strasse 2 64658 Fürth-Weschnitz Phone +49 6253 20080 © 2009 ProCredit Academy GmbH

F o r e w o r d P a g e | III

Foreword

Mathematics plays a crucial role in banks because all banking operations depend on accurate calculations and precisely defined methods. It is therefore not nearly enough to rely on our pocket calculator, our desktop PC or the server on our network. We need to develop an understanding of the operations and processes that are fundamental to all areas of our banks’ business. For this reason, a decision has been taken to introduce a Mathematics Training Programme at all of the banks and academies operated by ProCredit Holding. This Reader forms the basis for all of these courses.

The course material is divided into two main parts:

Vol. 1 comprises the absolutely fundamental basics of algebra. Confidence in applying these principles is essential. Numbers, variables, calculations, expressions, equations and simple functions are the basic components of mathematics and its applications. A bank cannot afford to show any weakness in any of these areas.

Vol. 2 addresses rather more advanced and in particular more bank-specific applications. In particular, it covers those functions that are used in the financial sector, as well as the main themes of financial mathematics, and finally the basics of statistics. This material will be taught in its full scope primarily at the Regional Academies; however, individual aspects will also be included in the bank-level training courses.

In line with this division into two parts, the material is also presented in two readers, the second of which builds upon the first. Accordingly, this second volume contains applications related to banking problems and some relevant topics of descriptive statistics.

The Reader is intended to define the subject matter to be covered during the training. It thus forms a shared platform for both trainers and course participants. For the training itself, the instructors are expected to produce an overview of the sections and pages to be worked on during each session. The overview, which should be handed out to the participants, could look something like this:

IV | P a g e F o r e w o r d

Block # Structure and Topics Pages

1 Volume 1 (revisited)

Introduction and Learning Targets

1 - 10

2 Special Functions

Exponential Functions

11 - 37

3 etc.

It is up to the trainer to select the sections that he or she will cover, depending on the prior knowledge of the participants and how quickly they are learning. It would therefore make no sense to try to offer a universal recommendation here. Experience to date has shown that it is better to cover the basics more slowly but therefore more thoroughly, rather than trying to pack as much as possible into a single course at all costs.

The main purpose of including references to specific sections and pages in the textbook is to allow the participants to use the Reader for their own independent study. The material has been developed very carefully and comprehensively, so that it can be followed without the help of a trainer. Numerous examples have been included in each section to illustrate the steps that have been described. All of the examples are highlighted in the text.

At the end of each section, we have set numerous practice exercises. These exercises are an absolutely indispensable part of any training course in mathematics. In fact, doing exercises is really the only way to learn mathematics. Students will always be able to follow a teacher who is good at motivating and explaining. But this is a very long way from saying that they will automatically be able, independently, to apply what they have understood. On the contrary: this ability only comes through independent practice with the exercises.

In the book, the exercises are immediately followed by an answers section, allowing the students to self-check their work. This section contains only the final answers, without explaining in detail how these results were arrived at. If you get stuck, please go back to the

F o r e w o r d P a g e | V

corresponding unit and try to find the reason why you failed. Studying means reading repeatedly.

Each chapter ends with progress tests designed to enable the students to monitor their own progress. Students should compare their answers against the ones given in the section immediately following the test. References to the sections of the Reader that are relevant to specific parts of the solution are designed to encourage students to go back and revise these sections in order to acquire the skills needed to solve problems of this particular type. However, the Reader does not contain any detailed explanations of the solutions to the progress tests.

We hope you enjoy working with this Reader.

VI | P a g e C o n t e n t s

Contents

The numbering of the chapters continues from the first reader, i.e. the first main chapter after description of the content of part 1 is the fifth chapter of the course material as a whole. The following sections are numbered accordingly.

1. Volume 1 (revisited) ......................................... 1 1.1 Abstract of Volume 1 ................................................. 2

1.2 Contents of Volume 1 ................................................ 6

1.3 Learning Targets ....................................................... 8

5. Special Functions ............................................ 11 5.1 Exponential Functions ............................................ 13

5.1.1 Graph of the Exponential Function ............................... 20

5.1.2 General Properties of Exponential Functions ................ 22

5.1.3 Application of the Exponential Function ...................... 23

5.1.4 Base e Exponential Function ......................................... 27

Exercise 5.1: Exponential Functions .................................. 32

Answers 5.1: Exponential Functions .................................. 34

5.1.5 Progress Test “Exponential Functions” ......................... 37

5.2 Logarithmic Functions ............................................ 39

5.2.1 Logarithmic Operations ................................................. 43

Exercise 5.2: Logarithmic Functions .................................. 46

Answers 5.2: Logarithmic Functions .................................. 47

5.2.2 Progress Test “Logarithmic Functions” ........................ 48

5.3 Answers to the Progress Tests ................................ 50

5.3.1 Answers to PT “Exponential Functions” ....................... 50

5.3.2 Answers to PT “Logarithmic Functions” ...................... 52

C o n t e n t s P a g e | VII

6. Time Value of Money ..................................... 55 6.1 Interest ..................................................................... 58

6.1.1 Simple Interest ............................................................... 60

6.1.2 Compound Interest ........................................................ 62

6.1.3 Effective Interest ........................................................... 69

Exercise 6.1: Interest ......................................................... 78

Answers 6.1: Interest .......................................................... 80

6.1.4 Progress Test “Interest” ................................................. 81

6.2 Deposits ................................................................... 83

6.2.1 Savings .......................................................................... 84

Single Payments ..................................................................... 84

Successive Payments .............................................................. 86

6.2.2 Regular Equal payments ................................................ 89

Exercise 6.2: Deposits ....................................................... 96

Answers 6.2: Deposits......................................................... 98

6.2.3 Progress Test “Deposits” ............................................... 99

6.3 Loans ..................................................................... 101

6.3.1 Repayments ................................................................. 102

Single Repayment ................................................................. 102

Several Repayments ............................................................. 103

Repayment Plan ................................................................... 104

6.3.2 Instalments ................................................................... 106

Outstanding Balance ........................................................... 112

6.3.3 Effective Interest Rate of Loans .................................. 114

Periodic Interest Rate .......................................................... 115

Repayment Plan and Effective Interest Rate ........................ 116

Floating Interest Rates......................................................... 120

Fees ...................................................................................... 122

VIII | P a g e C o n t e n t s

Exercise 6.3: Loans.......................................................... 129

Answers 6.3: Loans ........................................................... 132

6.3.4 Progress Test “Loans” ................................................. 136

6.4 Answers to Progress Tests .................................... 138

6.4.1 Answers to Progress Test “Interest” ............................ 138

6.4.2 Answers to Progress Test “Deposits” .......................... 139

6.4.3 Answers to Progress Test “Loans” ............................. 140

7. Basics of Statistics......................................... 143 7.0 Sum Symbol ........................................................... 147

Exercise 7.0: Sum Symbol ................................................. 152

Answers 7.0: Sum Symbol ................................................. 154

7.1 Presentation Techniques ....................................... 157

7.1.1 Tables .......................................................................... 158

7.1.2 Diagrams ...................................................................... 165

7.1.3 Sum Curves ................................................................. 168

7.1.4 Base Problem ............................................................... 171

Exercise 7.1: Presentation Techniques ............................. 175

Answers 7.1: Presentation Techniques ............................. 178

7.1.5 Progress Test “Presentation Techniques” .................... 185

7.2 Key Figures: Centre .............................................. 187

7.2.1 Mean ............................................................................ 189

7.2.2 Median ......................................................................... 196

7.2.3 Mode ............................................................................ 202

Exercise 7.2: Key Figures: Centre ................................... 205

Answers 7.2: Key Figures: Centre .................................... 208

7.2.4 Progress Test “Key Figures: Centre” .......................... 210

7.3 Key Figures: Spread ............................................. 212

C o n t e n t s P a g e | IX

7.3.1 Variance ....................................................................... 215

7.3.2 Standard Deviation ...................................................... 226

7.3.3 Normal Curve .............................................................. 230

Exercise 7.3: Key Figures: Spread ................................... 236

Answers 7.3: Key Figures: Spread ................................... 239

7.3.4 Progress Test “Key Figures: Spread” .......................... 241

7.4 Answers to Progress Tests .................................... 243

7.4.1 Answers to PT “Presentation Techniques” .................. 243

7.4.2 Answers to PT “Key Figures: Centre” ........................ 246

7.4.3 Answers to PT “Key Figures: Spread” ........................ 247

Index ........................................................................ 249

X | P a g e T a b l e s

Figures

Figure 5-1: Estimated population growth ........................................ 15

Figure 5-2: Growth function of water hyacinths .............................. 18

Figure 5-3: Different exponential functions with base a > 1 ........... 21

Figure 5-4: Different exponential functions with base a < 1 ........... 21

Figure 5-5: Inverse function of the exponential function ................. 40

Figure 6-1: Time line with relevant data .......................................... 57

Figure 6-2: Simple interest ............................................................... 61

Figure 6-3: Compound interest ......................................................... 63

Figure 6-4: Future value of money ................................................... 64

Figure 6-5: Present value of money .................................................. 66

Figure 6-6: Interest compounded p times per year ........................... 71

Figure 6-7: Compounding periodically ............................................ 73

Figure 6-8: Savings account ............................................................. 85

Figure 6-9: Successive payments ...................................................... 87

Figure 6-10: Positioning "today" ....................................................... 88

Figure 6-11: Regular equal payments ................................................ 90

Figure 6-12: Single repayment ......................................................... 103

Figure 6-13: Several repayments ...................................................... 104

Figure 6-14: Graph of the repayment plan ....................................... 106

Figure 6-15: Instalments ................................................................... 107

Figure 6-16: Calculation of outstanding amount ............................. 108

Figure 6-17: Floating yearly interest rates ...................................... 120

Figure 6-18: Floating periodical interest rates ................................ 121

Figure 6-19: Fees and one repayment .............................................. 123

Figure 6-20: Fees and instalments ................................................... 125

T a b l e s P a g e | XI

Figure 7-1: Column diagram of grades .......................................... 165

Figure 7-2: Column Diagram of a frequency table ........................ 166

Figure 7-3: Column diagram of a grouped frequency table ........... 168

Figure 7-4: Column diagram of the frequency sum of grades ........ 169

Figure 7-5: Frequency sum of loans as a column diagram ............ 170

Figure 7-6: Cumulative sum curve and distribution function ......... 171

Figure 7-7: Statistical figures in the Excel status bar .................... 191

Figure 7-8: Histogram balanced at the average ............................ 197

Figure 7-9: Right-tail distribution .................................................. 200

Figure 7-10: Left-tail distribution ..................................................... 201

Figure 7-11: The tails of frequency distributions ............................. 201

Figure 7-12: Narrowly spread attribute values ................................ 214

Figure 7-13: Widely spread attribute values .................................... 214

Figure 7-14: Spread around mean value .......................................... 228

Figure 7-15: Distribution of the sizes of newborn babies ................ 231

Figure 7-16: Histogram and fitted normal curve ............................. 232

Figure 7-17: The former DM 10 bill with normal murve ................. 233

Figure 7-18: A normal-shaped histogram ........................................ 234

XII | P a g e T a b l e s

Tables

Table 5.1: Estimated world population (in thousands) .................. 14

Table 5.2: Growth of water hyacinths ............................................ 17

Table 5.3: Area Covered by water hyacinths .................................. 18

Table 5.4: Limit process for number e ............................................ 29

Table 6.1: Future values ................................................................. 71

Table 6.2: Effective interest ............................................................ 72

Table 6.3: Effective interest rates ................................................... 76

Table 6.4: Repayment plan ........................................................... 105

Table 6.5: Repayment plan with monthly instalments .................. 117

Table 6.6: Repayment plan with quarterly instalments ................ 118

Table 6.7: Repayment plan with floating interest rate ................. 119

Table 7.1: List of grades ............................................................... 159

Table 7.2: Frequency table of grades ........................................... 159

Table 7.3: Frequencies of grouped data ....................................... 161

Table 7.4: List of data ................................................................... 163

Table 7.5: Grouped frequencies ................................................... 164

Table 7.6: Frequency density of grouped data ............................. 167

Table 7.7: “Base-Problem” .......................................................... 174

Table 7.8: Grouped frequencies (Copy of Tab. 7.5) ..................... 195

Table 7.9: List of 60 numbers ....................................................... 196

Table 7.10: Number list .................................................................. 215

Table 7.11: Calculating the variance ............................................. 216

Table 7.12: Variance of grouped data ............................................ 225

1 . V o l u m e 1 ( r e v i s i t e d ) P a g e | 1

1. Volume 1 (revisited)

Because of its sheer size, this course, which was originally conceived of as a single unit, has been divided into two parts and now appears in the form of two separate readers. The first part mainly concerned itself with the basics of algebra and simple functions. Most of this material is normally taught in high school, so the first part can be regarded as an introductory refresher course.

In contrast, this second part is devoted more to the applications of mathematics, though naturally our selection is biased towards those areas that play a role in everyday banking business. The mathematical foundations underlying credit operations form one focal point, descriptive statistics another.

Needless to say, this second part is based on the assumption that the student has understood the basics discussed in the first volume. In order to form a bridge from the first part and from earlier training courses that you may have attended, this first introductory chapter gives a very brief review of the topics covered in Volume 1. We have also included the table of contents for Volume 1 to make it easier for you to refer back to specific topics. Please note that, of course, the page numbers in this table of contents refer to the first volume of the Reader. However, the numbering of the chapters – with the exception of this first introductory chapter – follows on consecutively from the first volume. Hence, the next chapter is numbered 5 because the first volume ended with Chapter 4.

Finally, we re-iterate the training objectives, which naturally also apply to this part and thus to the course as a whole.

2 | P a g e 1 . V o l u m e 1 ( r e v i s i t e d )

1.1 Abstract of Volume 1 The introductory Chapter 1 of Volume 1 focused primarily on the role of mathematics as a language used to “translate real-world problems into models which can be processed and solved using mathematical methods”. With reference to examples, the transformation process from real problem to solution was described.

The array of mathematical instruments which are available for solving real-world problems are drawn initially from the basic instruments of algebra. They were covered in Chapter 2.

Algebra could be regarded as the area of mathematics that deals in the broadest sense of the term with “calculation”. At school we first learn to calculate with numbers, that is, to apply certain operations such as addition, subtraction, multiplication or division. Learning to calculate with whole, natural numbers usually presents no problem. However, people often start to make mistakes when they try to apply the same operations to fractions and decimals. Fractions and decimals play an important role in the banking sector, especially in connection with percentages. A strong command of fractions in all their variations is therefore indispensable.

That said, it would be inaccurate to equate algebra or indeed mathematics with “calculation”, just as it would be a ridiculous understatement to describe the literature of Shakespeare or Goethe as “letters”. One of the main characteristics of mathematics is that its arguments consist not only of numbers but also and above all of symbols, which stand for general parameters or unknown values. Parameters and variables are representatives of numbers. We can “calculate” with them in just the same way as we do with known values, and in particular we can use them to model real problems. It is only when we start to use variables in more complex mathematical expressions and in functions that mathematics becomes a powerful instrument that helps us generalise simple relationships. Variables and parameters allow us to take the decisive step from the specific to the general, and to analyse relationships that would otherwise appear too confusing.

In the calculation of compound interest, algebraic expressions in which numbers and variables are combined with exponents play an important role in banking-specific applications. A discussion of the basics would


therefore not be complete without a section devoted to the calculation rules that apply to integral and fractional exponents. As they are not always covered sufficiently intensively at school, revision of the methods for dealing with them are an important part of our course.

Through the combined application of mathematical operations on numbers and variables, we can form complex expressions (indeed, there is no limit to the complexity). One of the defining characteristics of mathematics is that when performing calculations with these expressions exactly the same rules must be applied as when calculating with numbers. We might sometimes tend to forget this when the expressions get really complicated. Yet it remains true to say: no matter how complicated the expression is, the rules of algebra always remain in force. In order to avoid confusion, however, it is usually necessary, and certainly recommended, to use brackets to remove any ambiguity from an expression. In other words, when solving expressions, it is better to use too many brackets rather than one set of brackets too few.

Equations are the most frequently used relations in mathematics. In mathematical models, i.e., in representations of real-world problems, they provide the framework. They are created when accounting equations are performed, physical laws are formulated or certain quantities for the model are defined. The first volume of the Reader therefore included a special section devoted mainly to explaining how equations are used in the modelling process.

If the modelling process gives rise to an equation, then the primary goal is usually to solve it. “Solve” means to determine the values of the variables in the equation that satisfy the conditions of that equation. Frequently the model will contain several equations, so-called systems of equations. In that case, the solutions must satisfy the conditions of all equations simultaneously. Ideally, they will then represent the solution to the original real-world problem.

Linear and quadratic equations are, on the one hand, the simplest forms of equations. At the same time, however, they are also the forms most frequently used in economics. They were discussed in detail in two sections of Chapter 3.

The fourth chapter then dealt for the first time with functions.

In our daily life we quite often encounter the verb “to function”. When we turn the ignition key and the engine of our car starts, we could say


that the starter motor is functioning correctly. When we switch on the light and it becomes bright in the room, we would agree that the switch is functioning correctly. Similarly, we could say that an organisation functions if everything goes smoothly. In all cases, we say “it functions” when an action leads to an expected and desired reaction.

If we want to describe the actions and reactions mathematically, starting a car or switching on a light are not very interesting examples because they are singular events. However, think about utility bills, in which the final amount depends directly on the usage. In this case the action is the consumption and the reaction is the amount shown on the invoice; thus, there exists an almost infinite number of possible pairings of:

Action → Reaction

Both sides of this relation can be described in terms of numbers. Hence, mathematics could serve very effectively as a language to describe the dependencies between consumption and bill.

In fact, this is one of the major concepts in mathematics: to describe, by means of equations, the relationships between factors which influence each other, and the rules which govern how they influence each other. An equation states the relationship between different variables. If we want to describe the reaction of a factor (given by a variable, say y) to the action of some other factor (expressed by another variable, say x), we could write:

In mathematics the relation between different variables is called a “function”. Functions form a key “vocabulary” in the language “mathematics”, because practically all dependencies on real problems must be described in terms of functions in order to apply mathematics to solve them. That is, a model basically consists of a set of functions. If we want to use models we have to know what to do with them.

Almost everything in mathematics is related to the concept of functions. However, that does not mean that we have to learn everything about them. Fortunately, we can easily restrict ourselves to

• the most basic functions,

• their representation,

Action xRule orRelation

Reaction y


• their properties, and

• their use.

Following the methodological approach of this whole reader, we began with the general properties of functions, which are more or less valid for all functions. In addition, we discussed general approaches to analysing functions based on a number of fundamental properties such as zero points, y-intercepts, slopes, and maximal or minimal points.

Very many functional relationships between economic quantities are linear in nature, meaning that the resultant quantity is directly proportional to the quantity that caused it. For example, sales revenues are proportional to the quantity and the price of the goods or services sold. Linear functions with one independent variable are represented graphically by straight lines on the Cartesian plane. Based on this simplest of all equations it is possible to explain the most important properties and analytical steps in a particularly descriptive way.

Another section of Volume 1 was devoted to a discussion of quadratic functions. In the case of a single independent variable, they are represented by parabola. As the simplest type of non-linear function, they provide a very descriptive example which can be referred to when discussing more complex functions, such as the exponential and logarithmic functions that are so important in banking. These special functions are covered in this volume in the next chapter, Chapter 5.


1.2 Contents of Volume 1 Volume 1 of the reader contained the basics of algebra and functions. The page numbers shown here refer to Volume 1.

1. Introduction ......................................................... 1

1.1 The Language Mathematics ...................................... 2

1.2 How to Apply Mathematics ....................................... 4

1.3 Learning Targets ....................................................... 8

2. Basic Algebra ..................................................... 11

2.1 Numbers .................................................................. 14

2.1.1 Numbers and Operations .............................................15

2.1.2 Fractions and Decimals................................................25

2.1.3 Percentages ..................................................................42

2.2 Exponents ................................................................ 58

2.2.1 Integer Exponents ........................................................61

2.2.2 Fractional Exponents ...................................................71

2.2.3 Radicals........................................................................80

2.3 Expressions.............................................................. 93

2.3.1 Integer Expressions ......................................................95

2.3.2 Fractional Expressions ...............................................104

C o n t e n t s o f V o l u m e 1 P a g e | 7

3. Equations .......................................................... 123

3.1 Use of Equations ................................................... 125

3.1.1 Modelling with Equations .........................................128

3.1.2 Solution ......................................................................132

3.2 Linear Equations ................................................... 147

3.2.1 Normal Form of Linear Equations .............................148

3.2.2 Solution ......................................................................149

3.3 Quadratic Equations ............................................. 158

3.3.1 Forms of Quadratic Equations ...................................159

3.3.2 Solution ......................................................................160

4. Basic Functions ................................................... 177

4.1 Properties of Functions ......................................... 184

4.1.1 Characteristics of the Graph ......................................185

4.1.2 Inverse Functions .......................................................189

4.2 Linear Functions ................................................... 203

4.2.1 Graph of a Linear Function........................................205

4.2.2 Properties of Linear Functions...................................208

4.3 Quadratic Functions ............................................. 216

4.3.1 Completion of the Square ..........................................218

4.3.2 Graph of the Quadratic Function ...............................220

4.3.3 Properties of the Quadratic Function .........................227


1.3 Learning Targets ProCredit is a bank. We aim to advise our customers, and above all we want to give them correct and transparent advice. In order to do so, a basic knowledge of algebra and some functions is indispensable.

Many of you will say: “But we’ve got computers and pocket calculators.”

That is a valid point, insofar as the software can ultimately calculate the exact amount down to the last cent. However, there are various sources of error: for example, you might enter the wrong figure. In this case, you should be capable of judging whether the result is plausible or not.

You might find yourself in a situation in which you have to advise someone and you don’t have your calculator or laptop with you. What would you say to your friend, for example, if he stopped you on the street and asked how much a loan would cost, or what the effective interest rate on a savings deposit would be?

In our daily work, with or without customers, with or without colleagues, with or without staff members: the essential details of banking business are based on handling quantitative variables, figures and operations. It is therefore essential to know

• the basics of algebra,

• some of the functions that are important for banking, and their properties,

• certain aspects of statistics, and

• the background to the deposit-taking and lending business.

That is why in this course we will start by learning a sufficient amount of vocabulary. We will familiarise ourselves with the absolutely fundamental concepts of algebra, such as numbers and operators, and in particular also fractions and exponents or indices, as well as how they are combined to form algebraic expressions.

In our daily lives and in our work, quantities are very often compared and weighed against each other. In mathematics, this generally leads to equations that one would like to solve.

1 . 3 L e a r n i n g T a r g e t s P a g e | 9

The most important form of an equation in mathematics is the function, through which causal relationships in the real world can be described in the mathematical model.

Basics of Algebra, Functions, and

Statistics

2. Basic Algebra 3. Equations

6. Time Value of Money

7. Basic Statistics

4. Basic Functions

5. Special Functions

1. Introduction

Vol

. 2: A

dvan

ced

In the bank, interest rates play an all-important role: calculating the return on equity requires us to focus particularly on exponents when working with algebraic expressions. Growth processes resulting from compound interest are described by means of so-called exponential functions, which naturally play an important role in describing problems in the field of financial mathematics.

Because of the return on capital, the value of capital is time-dependent, and because of this, financial-mathematical applications are often referred to as the “time value of money”. In these course units, which are


of direct relevance to applications in the bank, the indispensable fundamental principles of interest, deposits and loans will be described.

The above graphic shows the topics covered by this reader in their context. They are designed to be taught in an advanced course, which covers topics more closely related to applications in the banks.

The numbering of the chapters continues from the first reader, i.e. the first main chapter after this introduction is the fifth chapter of the course material as a whole. The following sections are numbered accordingly.

Each section contains numerous exercises. To allow you to check your results, the correct answers are given at the end of each chapter. In most cases, only the final answer is shown, but not the method of calculation in all its details. We deliberately chose not to show the recommended method because we wanted to encourage you to consider alternative methods if you get the wrong answer.

5 . S p e c i a l F u n c t i o n s P a g e | 11



Statistics



7. Basic Statistics

4. Basic Functions


1. Introduction

Vol

. 2: A

dvan

ced

Prerequisites: Functions are one of the most important fields of mathematics. In the first part of the course we learned about the general properties of functions, and focused on linear and quadratic functions. To follow the discussion on exponential functions, which are so important for the banking and financial sector, familiarity with exponents is a basic requirement.

Learning Targets: Exponential functions and their inverses, known as logarithmic functions, together play a major role in

12 | P a g e 5 . S p e c i a l F u n c t i o n s

the banking sector because they are needed to calculate interest charges on capital and evaluate cash flows.

All of the functions we have considered so far have been polynomial functions. In this reader we will discuss neither the rather large class of rational functions – those which contain the operation “division” – nor the functions containing roots. However, there is another class of functions which cannot be described by means of the algebraic operations addition, subtraction, multiplication, division, and taking of powers and roots on variables. These are called “transcendental” functions. Examples include the so-called trigonometric functions (sin, cos, tan, ctg), which are extremely useful for technical applications.

Given our interest in economics and especially in banking, we will define and investigate the properties of two other types of functions which belong to the “transcendental” class: exponential functions and logarithmic functions. They are used in describing and solving a wide variety of real-world problems, including the growth of populations of humans, animals, and bacteria; radioactive decay (negative growth); normal distribution in statistics (see: Chapter 7); the growth of money at compound interest; the absorption (negative growth) of light as it passes through air, water, or glass; and the magnitude of sounds and earthquakes. We will consider applications in these areas in more detail in the chapters that follow (6. Time Value of Money, and 7. Basic Statistics).

5 . 1 E x p o n e n t i a l F u n c t i o n s

5.1 Exponential Functions

Prerequisites: Exponential functions occur if the variable is part of the exponent. Therefore exponential expressions is a very basic for understanding these functions.

Learning Targets: In this chapter we define and investigate the properties of an important class of functions, called exponential functions. They are useand solving a wide variety of realespecially processes of growth and decay. With respect to bankingthe continuous compounding of interest.



Graph

Properties

5 . 1 E x p o n e n t i a l F u n c t i o n s P a g e | 13

Exponential Functions

Exponential functions occur if the variable is part of the exponent. Therefore confidence in handling exponential expressions is a very basic precondition for understanding these functions.

In this chapter we define and investigate the properties of an important class of functions, called exponential functions. They are used in describing and solving a wide variety of real-world problems, especially processes of growth and decay. With respect to banking, exponential functions describe

continuous compounding of interest.


5.2 Logarithmic Functions

Graph

Properties


Before we start to define some new mathematical concept we should spend some time looking at the growth of the world’s population. The following Tab. 5.1 shows the historical development of the estimated number of people living on Earth (see: Wikipedia).

Year Population Year Population Year Population

70,000 BC 2,000 1,000 BC 50,000 1960 2,981,659

10,000 BC 1,000 500 BC 100,000 1965 3,334,874

9,000 BC 3,000 1 200,000 1970 3,692,492

8,000 BC 5,000 1000 310,000 1975 4,068,109

7,000 BC 7,000 1750 791,000 1980 4,434,682

6,000 BC 10,000 1800 978,000 1985 4,830,979

5,000 BC 15,000 1850 1,262,000 1990 5,263,593

4,000 BC 20,000 1900 1,650,000 1995 5,674,380

3,000 BC 25,000 1950 2,518,629 2000 6,070,581

2,000 BC 35,000 1955 2,755,823 2005 6,453,628

Table 5.1: Estimated world population (in thousands)

If we look at the data in the table first, we notice that the world’s population grew very slowly up to the year 1800. Then, all of a sudden, the population started to grow rapidly.

The world population is the total number of humans living on Earth at a given time. As of September 2008, the world’s population is estimated to be just over 6.721 billion. In line with population projections, this figure continues to grow at rates that were unprecedented before the 20th century, even though the rate of growth has almost halved since its peak of 2.2% per year, which was reached in 1963. The world’s population, on its current growth trajectory, is expected to reach nearly 9 billion by the year 2042.

If we draw the graph of the function p = f(t) i.e. number of people as a function of time for the domain we get the picture shown in Fig. 5-1


(Source: Population Growth, Wikipedia). It shows a very distinct curve, which remained almost totally flat for a very long period. Apparently, the growth rate increased from the year 1000 on, then accelerated explosively from 1800 and especially in the 20th century.

200010002000 AD140006000800010000 BC0

1

2

3

4

5

6

Wor

ld P

opul

atio

n (B

illio

n)

Year

Figure 5-1: Estimated population growth

Let us look for additional examples. In the German newspaper “Frankfurter Rundschau”, dated November 10th, 2008, there was an article by A. Beutelspacher with the headline: “Out of control – Woe betide us if exponential functions start to gain momentum”. He described three situations.


1. Folding a newspaper exactly in the middle you get two layers of

half the area. Folding the smaller area again exactly in the middle, the result will be layers. If you continue to fold it in the same way, you get 8, then 16, then 32 layers.

Each step of the folding process doubles the number of layers and decreases the area by half.

Now, imagine that you can continue with the folding process as long as you want. Certainly, if you really try to do it you will come to an end pretty soon, because you will find that you physically cannot fold it any more. The area will be too small, even if you begin with a very large-sized newspaper.

But assuming that we could theoretically continue folding the pieces, we could pose the question: How often do I have to fold the newspaper until the stack of layers is so high that it closes the gap between the Earth and the moon, which is about 380,000 km?

Although this is a totally unrealistic problem, a mathematician is always allowed to ask questions like this because, as we will see, it is a nice example with which to illustrate the behaviour of processes which might get out of control.

Before reading on, just pause for a moment and guess how often the newspaper needs to be folded in order to reach the moon. The solution will be given at the end of this introductory section.

2. There is a famous story about the inventor of chess whose monarch wanted to reward him for this wonderful game. The monarch offered him one free wish. Without hesitating, the inventor answered that he would like the monarch to put one grain of rice on the first square of the chessboard, two on the second, four on the third, etc., doubling the number of grains each time until all of the squares had been filled.

It does not sound as though the total amount of rice would be very large, does it? However, taking into consideration that the chessboard has 64 squares, the number of grains of rice on the last

square alone would be which is almost the number 1 followed by 19 zeros. This is many times the total amount of rice grown on Earth in a year.

2 2 4⋅ =

632


3. While the first two examples are just amusing puzzles, the last

one is real and dramatic. Lake Victoria in East Africa, between Tanzania, Uganda and Kenya, is the second largest freshwater lake on Earth, and provides the local population with huge quantities of fish. Or at least, it used to – until the arrival of the water hyacinth. This pretty, decorative flower is in fact turning out to be a kind of monster because of the phenomenal speed at which it is spreading over the surface of the lake: when conditions are ideal for it, the area covered doubles every 15 days.

So it is just a question of time before the entire surface will be covered by an impenetrable carpet of plants which destroys the shore, blocks the waterways, and prevents the infiltration of oxygen, thus killing off the fish.

Recent explorations have shown that the carpet is growing at a rate of 2000 hectares per week. If you measure every week, you may find out that the growth per week depends on the area covered the week before. The rate might be, say, 4%. That means, if in one week the area covered is a then in the next week the area will be approximately:

0.04 (1 0.04) 1.04a a a a+ = ⋅ + = ⋅

During the following week, the covered area grows to:

21.4 (1 0.04) 1.04a a⋅ + = ⋅

because the rate of growth depends on the area covered in the previous time period. It is a natural consequence of organic growth due to cell division.

Starting the observation with the area a in week no. 0 you may get the figures in Tab. 5.2 for the following 8 weeks.

Week 0 1 2 … 7 8

Area 1.04a ⋅ 21.04a⋅ … 71.04a⋅ 81.04a⋅

Table 5.2: Growth of water hyacinths

a


The area covered is obviously a function of time t and it is certainly not too difficult to work out the structure of the

corresponding function ( ) 1.04ty f t a= = ⋅ where t denotes the number of weeks after the observations started.

Since a is simply a constant, it might be more interesting to study the function:

( ) 1.04ty f t= =

Using the following value table the graph of the above function is drawn in Fig. 5-2.

Year 0 1 2 3 4 4.3 4.6 5

Week 0 52 104 156 208 225 242 260

Area 1 7.7 54.6 419 3490 6800 13245 29022

Table 5.3: Area Covered by water hyacinths

0 1 year0

30000 a

Are

a co

vere

d (h

ecta

res)

20000 a

10000 a

5000 a

2 year 3 year 4 year 5 year

Figure 5-2: Growth function of water hyacinths


Comparing the two curves in figs. 5-1 and 5-2 we may discover similarities: Both functions are continuously increasing. In both cases, the rate of growth starts relatively slowly, and accelerates with increasing time. From a certain point on the gradient of the function becomes very steep – it goes “out of control”.

Both functions are very realistic examples of the exponential functions which we will study in this section.

We still have not answered one of the opening question: How do we have to fold the newspaper until the stack of paper reaches the moon? Folding

a newspaper ten times leads to a stack of 10 32 1024 10= ≈ pages. We can assume that this stack is about 10 cm high (comparing it with two reams of copy paper).

10 packs reach the height of 1 m → 410 pages

310 packs are 1 km high → 710 pages

The distance to the moon is about 384,000 km 54 10≈ ⋅ km; therefore we need:

5 7 124 10 10 4 10⋅ ⋅ = ⋅ pages

Now let’s calculate how many pages we produce by folding the

newspaper. Folded once, we get 12 2= pages; folding twice, we obtain 22 4= pages, folding it n-times, we obtain 2n pages.

The final question is now: For what value of n is 122 4 10n = ⋅ ?

The amazing result is that the above relation is true, when:

42 122 4.4 10≈ ⋅

Did your guess come close to that surprisingly small number of fold? Most probably you didn’t not because you likely didn’t take into account that you were confronted with the special properties of an exponential function.


5.1.1 Graph of the Exponential Function

Let us start by noting that the two functions f and g given by

and

are not the same function – indeed, they are not even similar or related to each other. The latter is called a power function, which is a special case of a polynomial function. The variable x is raised to the power of 2.

In the function ( ) 2xf x = the variable appears in the exponent. It is called the exponential function.

EXPONENTIAL FUNCTION

The equation:

with a > 0, a ≠ 1

defines an exponential function for each different a, called the base. The independent variable x appears in the exponent and may be any real value.

The domain of these functions is the set of real numbers. It can be shown that the range of all exponential functions is the set of all positive real numbers. The base is required to be positive to avoid imaginary numbers

such as .

Graphs of exponential functions for different bases a > 1 are shown in Fig. 5-3. From left to right the curves in the first quadrant show the functions:

( ) 5xy f x= = ; ( ) 2xy g x= = ; ( ) 1.2xy h x= =

Notice that, in the upper right (first) quadrant, the greater the bases of the function are, the steeper the curves are. In the upper left (fourth) quadrant however, exactly the opposite is true: The functions become flatter with increasing bases.

( ) 2xf x = 2( )g x x=

( ) xy f x a= =

12( 2) 2− = −


y

x

0-2-4-6-8-10-12-14 2 4 6 8 10 12 14

2

4

6

8

10

12

14

16

18

20

22

24

26

-2

-4

-6

( ) 5xy f x= =

( ) 2xy g x= =

( ) 1.2xy h x= =

Figure 5-3: Different exponential functions with base a > 1

y

x

0-2-4-6-8-10-12-14 2 4 6 8 10 12 14

2

4

6

8

10

12

14

16

18

20

22

24

26

-2

-4

-6

15

( )x

y f x= =

12( )

xy g x= =

11.2

( )x

y h x= =

Figure 5-4: Different exponential functions with base a < 1


The bases for the four functions shown in Fig. 5-3 were all greater than 1. If we now choose different bases less than 1 we will get curves like the ones in Fig. 5-4 for the exponential functions. Now the functions are shown in the fourth quadrant from right to left in this order:

15( )

xy f x= = ; 1

2( )x

y g x= = ; 11.2( )

xy h x= =

The graphs in Figs. 5-3 and 5-4 suggest the following important general properties of exponential functions, which we state without a proof.

5.1.2 General Properties of Exponential Functions

• The domain is the set of real numbers. • The range is the set of positive real numbers. • All graphs are continuous. • For all functions the y-intercept is y = 1. • The exponential functions have no zeros (see: point 2.). • The curves approach the x-axis on one side asymptotically, i.e. they

come infinitely close but never touch it.

• If a > 1, then increases as x increases.

• If a < 1, then decreases as x increases

• The exponential function is a one-to-one function.

EXPONENTIAL OPERATIONS

Exponential functions obey the familiar laws of exponents we discussed earlier for integer exponents (see: Volume 1; Section 2.2.1), which are also valid for rational (fractional) exponents.

xaxa


For exponential functions with the same base a > 0 and x,y real numbers the following rules hold:

•

•

• 0 1a =

• 1 xx

aa

−=

•

•

•

For exponential functions with different bases a,b > 0 and the same exponent x (real number) we obtain:

•

•

•

For examples, see Volume 1, Section 2.2.1 of the reader.

5.1.3 Application of the Exponential Function

We now consider three applications in which exponential functions are utilised for analytical purposes: population growth, radioactive decay, and compound interest.

x y x ya a a +⋅ =x

x yy

aa

a−

=

( )yx x ya a ⋅=

xnn xa a=

x ya a x y= ↔ =

( )x x xa b a b⋅ = ⋅

xx

x

a a

bb

=

, for 0x xa b x a b= ≠ → =


POPULATION GROWTH

Our first example involves the growth of populations of organisms, such as humans, animals, insects, and bacteria. Populations tend to grow exponentially due to cell division. This means that their growth depends at each point in time on the number of active cells. Assuming that in the growth phase all cells will double within a certain time – the doubling time – this parameter is a convenient and easily understood measure of the growth rate. It is the time taken for the population to exactly double.

Over shorter time periods t the population growth can be described by the model:

0( ) 2tdP f t P= = ⋅

where P is the population at time t, 0P is the population at time t = 0, and

d is the doubling time.

Note that when t = d:

The population is double the original, as would be expected.

EXAMPLE

Mexico has a population of around 100 million, and it is estimated that it doubles within 21 years. Assuming that the population grows constantly at the same rate, what will be the population

a) 15 years from now? b) 30 years from now?

We use the above model with the parameters: million and

.

a) million

b) 269.18 million

0 0( ) 2 2ddP f d P P= = ⋅ = ⋅

0 100P =21d =

1521( ) (15) 100 2 164.067P t P= = ⋅ =3021( ) (30) 100 2P t P= = ⋅ =


RADIOACTIVE DECAY

Our second example deals with radioactive decay, which can simply be regarded as a negative growth phenomenon. Decay means that if you start with a certain amount A of a radioactive isotope, the amount declines exponentially over time. The so-called half-life is a property of every radioactive isotope. It defines the rate of decay in terms of the time taken for the material to decay to half of its original amount.

Over any other time period t the decay can be described by the model:

where A is the amount at time t, 0A is the amount at time t = 0, and h is

the half-life.

Note that when t = h:

The amount is half the original material, as would be expected.

EXAMPLE

The radioactive isotope gallium-67 has a half-life of 46.5 hours. If we start with 100 milligrams of the isotope, how many milligrams will be left after

a) 24 hours? b) 1 week?

We use the above model of radioactive decay with the parameters:

0 100A = mg; h = 46.5 and 1 week = hours.

a) ( )24

46.512( 24) 100 69.925A f t= = = ⋅ = milligrams

b) ( )16846.51

2( 168) 100 8.173A f t= = = ⋅ = milligrams

( )10 2( )

thA f t A= = ⋅

( )1 10 02 2( )

hhA f h A A= = ⋅ = ⋅

7 24 168⋅ =


COMPOUDED INTEREST

The third application is directly connected with your work in the ProCredit Bank and is absolutely fundamental to many topics in banking and finance. It relates to interest, which is generally calculated as a percentage of the money saved or borrowed (principal) over a period of time. Most often the percentage is the amount due for one year, but, as you are all aware, the interest rate can also be defined for other periods. We will discuss many topics relating to interest in much greater detail in the sixth chapter, “Time Value of Money” (see: Page 55). However, for the time being, let us simply consider the basic idea of interest and why it is a nice example of the use of exponential functions.

If at the end of an interest period the interest earned on a savings deposit is added to the balance, then in the next period interest is earned not only on the original deposit but also on the interest earned during the previous period. In other words, interest is paid on interest. This is called compound interest.

Assume the interest rate is i per year, compounded annually. This means that interest is capitalised (added to the balance) at the end of the year. The balance of a deposit D after one year is:

In the second year, interest is payable on the balance :

If we let be the balances after 1, 2, 3, 4, … years,

respectively, then the following model describes the balance after n interest periods (years):

where:

= balance after n interest periods

D = deposit at period n = 0

i = interest rate per interest period (year)

q = 1+i = interest factor (balance plus interest)

1 (1 )B D i D D i= + ⋅ = ⋅ +

1B

22 1 1 1 (1 ) (1 )B B i B B i D i= + ⋅ = ⋅ + = ⋅ +

1 2 3 4, , , ,...B B B B

(1 )n nnB D i D q= ⋅ + = ⋅

nB


EXAMPLE

Suppose you deposit D = EUR 1000 in a savings account that pays i = 6% interest, compounded annually.

a) How much will the balance be after 10 years?

D = 1000; i = 0.06; q = 1.06; n = 10; 10 ?B =

€

b) If you want the balance to have doubled 10 years from now, how high must the interest rate be?

D = 1000; i = 0.06; n = 10; 10 2000B = ; q = ?

From the above model we obtain:

(1 ) (1 ) 1n n n nnn

B BB P i i i

P P= ⋅ + → + = → + =

1nnB

iP

= −

110102 1 2 1 1.07177 1 0.07177i = − = − = − =

To double a principal in 10 interest periods, the interest must be i = 7.177%.

5.1.4 Base e Exponential Function

When introducing numbers we mentioned in Volume 1, Section 2.1.1 the three most commonly used irrational numbers: 2 , π , and e. The constant is a so-called natural constant which is well known from the

formulas used to calculate the circumference ( ) and the area ( ) of a circle (with radius r). The third irrational constant is at least as important as the number π. It is the number e = 2.71828182846… which is quite frequently the base of an exponential function describing many natural growth phenomena. The irrational number e is named after the great Swiss mathematician Leonard Euler (1707 – 1783).

10 1010 1000 1.06 1790.85B D q= ⋅ = ⋅ =

π2 rπ ⋅ 2rπ ⋅


The constant e occurs naturally when we study compound interest. Although we are going to study the effects of interest in more detail in the sixth chapter – as already mentioned – it is a good opportunity to discuss some aspects of compound interest in connection with the so-called e-function, which is a synonym for the exponential function with base e:

( ) xy f x e= = where e = 2.71828...

In the model of compound interest:

we assumed that interest is paid (compounded) annually. If interest i (the so-called nominal yearly rate) is compounded quarterly, monthly, or even daily, the frequency of capitalisation is 4, 12, or even 360 times, respectively, within one year. The corresponding formulas are:

• when compounding quarterly

• when compounding monthly

• when compounding daily

Generalising the formulas for compounding p times within a year we obtain:

• when compounding p times.

The balance after t years will be:

•

Substituting provides a slightly different form of

the same formula:

•

(1 )n nnB D i D q= ⋅ + = ⋅

( )41 41 iB D= ⋅ +

( )121 12

1 iB D= ⋅ +

( )3601 3601 iB D= ⋅ +

( )1 1p

ip

B D= ⋅ +

( ) ( )1 1tp p t

i it p p

B D D⋅ = ⋅ + = ⋅ +

1ip x

p i x= → = ⋅


• ( ) ( ) ( )1 11 1 1i tp t x i t xi

t p x xB D D D⋅⋅ ⋅ ⋅ = ⋅ + = ⋅ + = ⋅ +

So far we have only done some simple algebraic manipulations.

However, if we look more carefully at the expression within the square brackets we will realise that it tends to approach a limit.

x

1 2

10 2.59374246

50 2.69158880

100 2.704813829

500 2.715568520

1000 2.716923932

5000 2.718010050

10000 2.71814592

100000 2.71826824

Table 5.4: Limit process for number e

Although this series of numbers is not a proof, it is very obviously approaching the irrational number e. Mathematicians use the limit notation for describing approaches like this:

( )1lim 1 2.7182818...x

xxe

→∞+ = =

From the substitution we should recognise that increasing x means that the number of compounding periods increases equally. Therefore the limiting procedure simply means that we divide the year into an increasing number of interest periods. Hence describes

infinitely many periods, which means continuous compounding. The deposit is compounded every moment of time.

1( ) 1

x

f xx

= +

p i x= ⋅

limx→∞


CONTINUOUS COMPOUND INTEREST

If a deposit D is compounded continuously at a yearly rate of i, in nominal terms, then the balance after the time t is given by:

EXAMPLE

A deposit of EUR 1000 will be compounded for t = 5 years at a nominal interest rate i = 6%:

a) annually b) quarterly c) monthly d) daily

e) continuously

How high will the balances be in each case?

a) p = 1: EUR

b) p = 4: EUR

c) p = 12: EUR

d) p = 360: EUR

e) p = ∞: EUR

tB

( ) i ttB f t D e⋅= = ⋅

55 1000 (1 0.06) 1338.23B = ⋅ + =

200.065 41000 (1 ) 1346.86B = ⋅ + =

600.065 121000 (1 ) 1348.85B = ⋅ + =

18000.065 3601000 (1 ) 1349.83B = ⋅ + =

0.065 0.0655 1000 1000 1349.86B e e⋅ ⋅= ⋅ = ⋅ =


EXERCISE 5.1: EXPONENTIAL FUNCTIONS

You will find the answers to the exercise on the pages directly following the problems. They consist only of the final answer, allowing you to compare your answer with ours.

We deliberately chose not to show the recommended method because we wanted to encourage you to consider alternative methods if you get the wrong answer.

1. Sketch the following functions:

a) b)

c) d)

2. Simplify the following expressions:

a) b)

c) d)

e) f)

3. Which values solve the following equations?

a) b)

c) d)

e)

( ) 3xf x = ( )13

( )x

g x =

( ) (2 )xh x = − ( ) 2 xk x −=

5 743 7 10a a a⋅ ⋅

12

13

a

a

3 25 3x x x⋅ ⋅

122(9 )a

4 1 2 33 3x x+ −⋅ ( ) ( )22 23 3

a b

b a⋅

3 3 12 2x x− += 2 5 310 10 0x x− −− =2 1

1 3 22

55 5

5

xx x

x

−− −

−⋅ = 2 1 2 44 2x x− +=

4 2(2 1) 9x + =


4. The capital D = EUR 2000 was deposited at the beginning of

2004. What will be the balance at the end of 2007 at the nominal interest rate i = 5% when

a) compounded annually?

b) compounded monthly?

c) compounded continuously?

5. A country has a population of 28 million. During the past 17 years the population has doubled. What will be the estimated population

a) 10 years from now? b) 20 years from now?

c) 30 years from now?

6. A radioactive isotope has a half-life of 7 hours. How much of 25 milligrams will be left:

a) after 36 minutes? b) after 3 hours?

c) after 2 days?

d) How much was there 22 hours ago?


ANSWERS 5.1: EXPONENTIAL FUNCTIONS

1. a)

b) ( )13( )

xg x =

( ) 3xf x =y

x

0-1-2-3-4-5-6-7-8-9 1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8

-1

-2

-3

-4

-5

y

x

0-1-2-3-4-5-6-7-8-9 1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8

-1

-2

-3

-4

-5


c) ( )( ) 2xh x = −

d)

y

x

0-1-2-3-4-5-6-7-8-9 1 2 3 4 5 6 7 8

1

2

-1

-2

-3

-4

-5

-6

-7

-8

-9

-10

-11

( ) 2 xk x −=y

x

0-1-2-3-4-5-6-7-8-9 1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8

9

10

11

12

-1

-2


2. a) 577210a b)

16a c)

3415x d) 3a

e) f)

3. a) x = –2 b) x = 2 c) 12x = d) x = 3

e)

4. a) EUR 2431.01 b) EUR 2441.79 c) EUR 2442.81

5. a) 42.1 million b) 63.3 million c) 95.1 million

6. a) 23.56 milligrams b) 18.57 milligrams

c) 0.2156 milligrams d) 220.82 milligrams

33x+2

2

2

3

a b

b a

+

+

1x =


5.1.5 Progress Test “Exponential Functions”

You should assign yourself some time for concentrated work on this test. Try to solve as many problems as you can. Don’t use the reader to look for the solution. The aim of the test is to give you feedback on how much you know or have learned up to now.

At the end of the chapter you will find the solutions to the problems. Each solution is reduced to the final answer; it may even be only a number, a symbol, a table or a graph. You should check your answers. If they are correct, you may continue and start the next chapter. Otherwise (if you got the wrong answer or no answer at all) you will be given advice on what to do. In most cases you will be directed to the section in the reader which you should repeat in order to close the gap.

1. Sketch:

a) ( ) 2 2xy f x= = +

b) ( )14

xy =

c) ( )2x

y = −

2. Simplify the following expressions:

a) 2132x x

−⋅ = b)

54

17

x

x=

c) 2 1 2 3

3 2

4 4

4

x x

x

− +

−⋅ =

3. Which values solve the following equations?

a) 2 2 33 3x x− += b) 2 1

12( 2)

416

2

xx

x

−−

− =


c) 1 1xe − =

4. At the beginning of the year EUR 400 was deposited in an account paying 2% interest, compounded monthly. How much is in the account at the end of the year?

5. A radioactive isotope has a half-life of 16 hours. How much of 124 mg will be left after

a) 24 hours? b) 48 hours? c) 1 week?

6. A radioactive isotope has a half-life of 8 hours.

a) How much was there 24 hours ago, if there is 45 mg left?

b) After which time will 10 mg of 50 mg be left?

5 . 2 L o g a r i t h m i c F u n c t i o n s


Prerequisites: Since the logarithmic function is defined as the inverse function to the exponential functionhelpful to rewhich were explainedespecially the discussion of the inverse function.

Learning Targets: The discussion of the logarithmic function is very strongly connected to the discussion of its inverse function, thefunctions are onewill discuss how the general concept of inverse functions helps to define and understand the properties of logarithmic functions.

5. Advanced Functions


Graph

Properties

5 . 2 L o g a r i t h m i c F u n c t i o n s P a g e | 39

Logarithmic Functions

Since the logarithmic function is defined as the inverse function to the exponential function, it is

view the general properties of functions, were explained in Volume 1, Section 4.1, and

especially the discussion of the inverse function.

The discussion of the logarithmic function is very strongly connected to the discussion of its inverse

, the exponential function. Since exponential functions are one-to-one, their inverse exists. We will discuss how the general concept of inverse functions helps to define and understand the properties of logarithmic functions.

5. Advanced Functions


Graph

Properties


In Volume 1, Section 4.1, we extensively discussed one-to-one functions and the existence of the inverses to one-to-one functions. In addition we discussed a procedure for determining the inverse function. However, this procedure is only applicable to functions whose variables can be isolated.

The exponential function is clearly a one-to-one function. It is relatively easy to construct its inverse as a mirror at the 45° line: y = x (see: Fig. 5-5).

y

x

0-2-4-6-8 2 4 6 8 10

2

4

6

8

10

12

14

16

-2

-4

( ) xy f x a= =

1( ) ( ) logay g x f x x−= = =

Figure 5-5: Inverse function of the exponential function

However, there is no algebraic procedure for solving the exponential

function for x in order to determine the function

As early as the 8th century an Indian mathematician described values for the inverse functions of base 3 and 4. In the 13th century the first tables were produced by Arabian mathematicians and in 1614 the Scottish

( ) xf x a=

( ) xf x a= 1( ) ?f x− =


mathematician John Napier (1550 – 1617) published a book which made him the “inventor of the logarithm”, although the Swiss watchmaker Jost Bürgi (1552-1632) also independently developed a system for calculating logarithms. Hence, the name “logarithm” was created about 400 years ago to describe what is simply the inverse to the exponential function.

LOGARITHMIC FUNCTION

The inverse function of the exponential function with a > 0 is called the logarithmic function for base a and is denoted by:

where a > 0

The inverse function of the exponential function , i.e. to base e, is called the natural logarithmic function and is denoted by:

lny x=

The inverse function of the exponential function 10xy = , i.e. to base 10, is called the common logarithmic function (also decade logarithm) and is denoted by:

logy x= (sometimes also: lgy x= )

Although the world had been given a new function, there was no algebraic representation of it. Until the rise of electronic calculators, the only way to use the function was by means of the graph and table values. In fact, until the late 1960s logarithmic tables were an important part of school textbooks and technical libraries.

DEFINITION OF THE LOGARITHMIC FUNCTION

As the inverse of the exponential function, the logarithmic function is defined by:

log yay x x a= ↔ = with 0a > and 1a≠

xy a=

logay x=

xy e=


This means that the log to the base a is the exponent to which a must be raised to obtain x. Or in other words, the log reverses the exponential, i.e. executed consecutively they cancel each other out:

and

and

and

For some very obvious exponential numbers it is easy to deduce the logarithm:

2log 8 3= because 32 8=

3log 81 4= because 43 81=

log100000 5= because 510 100000= (Remember: is the

common logarithm.)

EXAMPLES

The following relationships between exponential functions and logarithmic functions hold:

1.

2.

3.

4.

5.

6.

log xa a x= loga xa x=

log10x x= log10 x x=

ln xe x= ln xa x=

10log log=

522 32 log 32 5= ↔ =

33.43.4 39.304 log 39.304 3= ↔ =

12 1

16 216 16 4 log 4= = ↔ =

34log 64 3 4 64= ↔ =

1.94285log 22.8 1.9428 5 22.8= ↔ =

0.22937log 0.64 0.2293 7 0.64−= − ↔ =


Similar to the properties of exponential functions (see: Section 5.1.2; Page 22) there are a set of rules for logarithms which are the exact inversions of the rules for exponential functions.

5.2.1 Logarithmic Operations

For logarithmic functions with the same base a>0 and x,y real numbers the following rules hold:

•

•

•

•

• log ( ) logya ax y x= ⋅

•

•

For logarithmic functions with different bases a,b > 0 and the same exponent x (real number) we get:

• ( )log ( ) log ( ) log logx x xa a a ay z y z x y z⋅ = ⋅ = ⋅ +

• ( )log log log logxx

a a a ax

y yx y z

zz

= = ⋅ −

• log loga bx x a b= ↔ =

log ( ) log loga a ax y x y⋅ = +

log log loga a ax

x yy

= −

log 1 0a =

1log log 1 log loga a a ax x

x = − = −

11log log lognn

a a anx x x= = ⋅

log loga ax y x y= ↔ =


EXAMPLES

The operations are illustrated by means of the following examples where the logarithms are finally determined with a calculator. We use the common logarithm; the formulas are the same for other logarithms with different bases.

1.

2. log 47849 log(4.7849 10000) log 4.7849 log10000= ⋅ = +

log 47849 0.67987 4 4.67987= + =

3. 3.531000

log 0.00353 log log3.53 log1000 0.54777 3= = − = −

log 0.00353 2.45223= −

4.

5.

If you look at a scientific electronic calculator you will notice that there are (usually) only two logarithmic functions, namely the natural logarithm and the common (decade) logarithm (log x is sometimes also denoted by lg x). How can we calculate the general logarithm to every base a > 0?

Going back to the definition and applying the above operations will help us:

with

Applying some known logarithm (for instance the natural logarithm) to both sides of the equation gives:

or, if you prefer to use the common logarithm:

( )log 200 log 2 100 log 2 log100 0.30103 2 2.30103= ⋅ = + = + =

( )0.68log 23.4 0.68 log 23.4 0.68 1.3922 0.93107= ⋅ = ⋅ =

( )0.2log 4.6 0.2 log 4.6 0.2 0.66276 0.13255− = − ⋅ = − ⋅ = −

(ln )x

log yay x a x= ↔ = 0 and 1a a> ≠

( ) lnln ln ln ln

lny x

a x y a x ya

= → ⋅ = → =

( ) loglog log log log

logy x

a x y a x ya

= → ⋅ = → =


EXAMPLES

Again, the final logarithms must be determined with a calculator:

1.

2. 0.34log16.9 1.2279

log 16.9 2.6208log0.34 0.4685

= = = −−

↔ 2.62080.34 16.9− =

3.

3.36245.7

ln 349 5.8522log 349 3.3624 5.7 349

ln 5.7 1.7405= = = ↔ =

66 8.584 6

5log10 6

log 10 8.584 5 10log5 0.699

= = = ↔ =


EXERCISE 5.2: LOGARITHMIC FUNCTIONS

You will find the answers to the exercise on the pages directly following the problems. They consist only of the final answer, allowing you to compare your answer with ours.

We deliberately chose not to show the recommended method because we wanted to encourage you to consider alternative methods if you get the wrong answer.

1. Determine x without using a calculator:

a) b) c) 0.5log 8x =

d) e) log 0.01x =

2. Solve the equations for x:

a) b) c) log 4x =

d) e)

3. Determine the unknown x in the following expressions:

a) b) c) log 256 4x =

d) e) ( )125 5

log x=

4. Use your calculator to finally solve the problems:

a) log 2.4x = b) log 0.02x = c) ln 0.00002x =

d)

5. Use the logarithm to solve the following equations:

a) b) c)

3log 81x = 1.1log 1.21x =

25log 5x =

2log 5x = 3log 0.5x =

4log 2x = 5log 0x =

4log 3x = 2log 8 x=

6log10 x− =

12log(2.4 10 )x = ⋅

100xe = 3 110 200x+ = 32.4 0.45x =


ANSWERS 5.2: LOGARITHMIC FUNCTIONS

1. a) b) c)

d) e)

2. a) b) c)

d) e) x = 1

3. a) b) c)

d) e)

4. a)

b)

c)

d)

5. a) b) c)

4x = 2x = 3x = −12

x = 2x = −

32x = 3x = 10000x =

16x =

64x = 3x = 4x =

6x = − 0.2153−

log 2.4 0.3802x = =

log 0.02 1.699x = = −

10.8198x = −

12.3802x =

4.6052x = 0.4337x = 0.3040x = −


5.2.2 Progress Test “Logarithmic Functions”

You should assign yourself some time for concentrated work on this test. Try to solve as many problems as you can. Do not use the reader to look for the solution. The aim of the test is to give you feedback on how much you know or have learned up to now.

At the end of the chapter you will find the solutions to the problems. Each solution is reduced to the final answer; it may even be only a number, a symbol, a table or a graph. You should check your answers. If they are correct, you may continue and start the next chapter. Otherwise (if you got the wrong answer or no answer at all) you should repeat the corresponding units in order to close the gap.

1. Sketch the logarithmic functions by using the property that it is the inverse to a certain exponential function:

a) 2( ) logy f x x= = is the inverse of: 2xy =

b) 0.5( ) logy g x x= = is the inverse of ( )12

xy =

c) ( ) ln 2y h x x= = is the inverse of 12

xy e=

2. Apply the calculation rule for logarithms to find simplified ways of performing these calculations:

a) 2log( )yx ⋅ b) logn

m

x

y

c) 2 2ln xe −

3. Calculate without a calculator:

a) 5log 125 b) 4.5lne c) 30.1log 10

4. Solve the equations for x:

a) 2log 16x = b) 5log 10x = c) 2.5log 0x =

P r o g r e s s T e s t P a g e | 49

5. Determine the unknown x in the following expressions:

a) 3log 1x = b) 3log10 x− = c) ln 2 1x =

6. Use the logarithm to solve the following equations:

a) 2 40xe = b) 2 110 1000x− = c) 3 12 128x+ =

7. Use your calculator to find:

a) 3.4log 57 b) 25.6log 12.9 c) 7.45log 0.94

d) 0.45log 6.77 e) 0.33log 0.79


5.3 Answers to the Progress Tests

5.3.1 Answers to PT “Exponential Functions”

You should check your answers. If they are correct, you may continue and start the next chapter. If not (if you got the wrong answer or no answer at all) repeat the topics of the corresponding sections in order to close the gap.

1. a) ( ) 2 2xy f x= = +

b) ( )14

xy =

c) Not defined

5 . 3 A n s w e r s t o P r o g r e s s T e s t s P a g e | 51

2. a) 2 113 62x x x

− −⋅ = b)

531428

17

xx

x=

c) 2 1 2 3

2 33 2

4 44

4

x xx

x

− ++

−⋅ =

3. a) x = −5 b) x = 3 c) x = 1

4. EUR 507.30

5. a) 43.84 mg b) 15.5 mg c) 0.0856 mg

6. a) 360 mg b) 18.57 hours


5.3.2 Answers to PT “Logarithmic Functions”

You should check your answers. If they are correct, you may continue and start the next chapter. If not (if you got the wrong answer or no answer at all) repeat the topics of the corresponding sections in order to close the gap.

1. a) 2( ) logy f x x= = is the inverse of 2xy =

2( ) logy f x x= =

2xy =

b) 0.5( ) logy g x x= = is the inverse of ( )12

xy =

0.5( ) logy g x x= =

( )12

xy =


c) ( ) ln 2y h x x= = is the inverse of 12

xy e=

( ) ln 2y h x x= =

12

xy e=

2. a) 2log( ) log log 2 logyx x y⋅ = + −

b) log log logn

m

xn x m y

y

= ⋅ − ⋅

c) 2 2 2ln 2xe x− = −

3. a) 5log 125 3= b) 4.5ln 4.5e = c) 30.1log 10 3= −

4. a) x = 65,536 b) x = 9,765,625 c) not feasible

5. a) x = 3 b) x = −3 c) x = 1.359141

6. a) x = 1.844 b) x = 2 c) x = 2

7. a) 3.4log 57 3.3038= b) 25.6log 12.9 0.7877=

c) 7.45log 0.94 0.00308= − d) 0.45log 6.77 2.395= −

e) 0.33log 0.79 0.2126=

6 . T i m e V a l u e o f M o n e y P a g e | 55



Statistics



7. Basic Statistics

4. Basic Functions


1. Introduction

Vol

. 2: A

dvan

ced

Prerequisites: It is required that one can deal competently with equations and functions. In particular, exponential and logarithmic functions will be applied.

Learning Targets: In order to understand the procedures in a bank, it must be understood that the value of capital always depends on the variable of time. In every calculation of cash flows, one must be able to determine and understand the value of time.

56 | P a g e 6 . T i m e V a l u e o f M o n e y

Some of the most fundamental roles of a bank are: collecting deposits, managing accounts and providing loans. All these processes involve charging interest on capital. This chapter will cover the basics of the associated calculations.

Today, most calculations are performed with computers. However, although it is possible to make such calculations with computers, accurate results depend on the correct entry of the figures. Therefore, it is important to develop a feeling for the plausibility of the calculated results. Only then is it possible to be sure that accurate and meaningful information is communicated to the customer.

The value of money is dependent on time. This implies neither that the value printed on the bill changes, nor that the value decreases due to inflation, but rather that the value changes when the money is invested. The main concern is how the value of investments (in-payments) and loans are affected by interest rates at different times. If we have an investment at a certain time and repayment at another time, the values of these two payments differ because of the possibility of earning interest. In a situation in which all cash flows are certain – which we will assume throughout this chapter – the rate of interest can be used to express the "time value of money. The rate of interest will allow us to adjust the value of cash flows, whenever they occur, at a particular point in time. It will thus be necessary to position time-adjusted cash flows at a single point in time.

A convenient way to represent cash flows at several points in time is to use a time line. It is always very helpful to think of the time line as some imaginary account where, over time, we can mark all relevant payments with arrows (see: Fig. 6-1). Throughout this course we will represent in-payments with arrows towards the time line, indicating that there is cash flow into the account, and mark out-payments with arrows pointing downwards from the timeline, indicating that there is a cash flow out of the account. Points in time (dates) are represented by points on the scale.

Since cash flow changes with time, we can only add, subtract, compare, and balance cash flows when all payments are adjusted to a single point

6 . T i m e V a l u e o f M o n e y P a g e | 57

in time. This point is usually marked extra as the reference point or often simply by the word “today”.

t (time line)

in-paymenti.e. D = 200 EUR

out-paymenti.e. W = 100 EUR

time period reference point

interest rate, i.e. i = 5%

today

Balance

discountingcompounding

compounding

datei.e. 5.5.2009

Figure 6-1: Time line with relevant data

The time line already contains all the relevant information. We will make extensive use of it to structure real-world situations graphically. Once you have configured a graphical model with all the relevant information about a problem, it is much easier to translate it into the “language” of mathematics, and then solve it using known mathematical procedures.

58 | P a g e

6.1 Interest

Prerequisites: Interest is the reason for the growth of money. If interest is paid exponential expressioninterest procedures.

Learning Targets: Interest is the reason why the value of money depends on the time. This understanding financial operationsbasis of banking. We will discuss various aspects of interest, like simple and compound interest. In order to compare retail producteffective interest is essential.


6.1 Interest

Simple Interest

Compound Interest

Effective Interest

6.2 Deposits

Savings

6 . T i m e V a l u e o f M o n e y

Interest is the reason for the growth of money. If interest is paid on interest, then we must use exponential expressions, an indispensible part of interest procedures.

Interest is the reason why the value of money depends on the time. This key fact is fundamental in understanding financial operations, and therefore the basis of banking. We will discuss various aspects of

like simple and compound interest. In order to compare retail products, an understanding of effective interest is essential.


6.2 Deposits

Savings

Saving Rates

6.3 Loans

Repayments

Instalments

Effective Interest

6 . 1 I n t e r e s t P a g e | 59

INTEREST IS THE PRICE PAID FOR THE USE OF SAVINGS OVER A GIVEN

PERIOD OF TIME .

From Wikipedia:

"In the Middle Ages, time was considered to be property of God. Therefore, to charge interest was considered to commerce with God's property. Also, St. Thomas Aquinas, the leading theologian of the Catholic Church, argued that the charging of interest is wrong because it amounts to "double charging", charging for both the thing and the use of the thing. The church regarded this as a sin of usury; nevertheless, this rule was never strictly obeyed and eroded gradually until it disappeared during the industrial revolution.

Usury has always been viewed negatively by the Roman Catholic Church. The Second Lateran Council condemned any repayment of a debt with more money than was originally loaned, the Council of Vienna explicitly prohibited usury and declared any legislation tolerant of usury to be heretical, and the first scholastics reproved the charging of interest. In the medieval economy, loans were entirely a consequence of necessity (bad harvests, fire in a workplace) and, under those conditions, it was considered morally reproachable to charge interest.

In the Renaissance era, greater mobility of people facilitated an increase in commerce and the appearance of appropriate conditions for entrepreneurs to start new, lucrative businesses. Given that borrowed money was no longer strictly for consumption but for production as well, it could not be viewed in the same manner. The School of Salamanca elaborated various reasons that justified the charging of interest. The person who received a loan benefited and one could consider interest as a premium paid for the risk taken by the loaning party. There was also the question of opportunity cost, in that the loaning party lost other possibilities of utilizing the loaned money. Finally and perhaps most originally was the consideration of money itself as merchandise, and the use of one's money as something for which one should receive a benefit in the form of interest. Martín de Azpilcueta also considered the effect of time. Other things being equal, one would prefer to receive a given good now rather than in the future. This preference indicates greater value. Interest, under this theory, is the payment for the time the loaning individual is deprived of the money.


Economically, the interest rate is the cost of capital and is subject to the laws of supply and demand of the money supply. The first attempt to control interest rates through manipulation of the money supply was made by the French Central Bank in 1847.

The first formal studies of interest rates and their impact on society were conducted by Adam Smith, Jeremy Bentham, and Mirabeau during the birth of classic economic thought. In the early 20th century, Irving Fisher made a major breakthrough in the economic analysis of interest rates by distinguishing nominal interest from real interest."

6.1.1 Simple Interest

Simple interest is interest that is paid (earned) on only the original amount, or principal, that was borrowed (lent). The amount (in euro) of simple interest is a function of three variables:

1. the original amount borrowed (lent), or principal ,

2. the interest rate per interest period, and

3. the number of periods for which the principal is borrowed (lent).

The interest period is the time which elapses between successive dates at which interest payments are due. Often, the interest period is one year. However, we are going to discuss shorter periods as well, e.g., months or days.

Simple interest is interest that is paid (earned) on the principal only. For the deposit D, the interest earned at a rate of i after one period is:

I D i= ⋅

6 . 1 I n t e r e s t P a g e | 61

t

interest rate i

.....I D i= ⋅ I D i= ⋅ I D i= ⋅

D

Figure 6-2: Simple interest

After n periods the total interest paid is:

The balance of the account is still D. In Germany it is possible to buy certain types of bonds which are provided with simple interest.

EXAMPLE

For the deposit D = 200 EUR and an interest rate of i = 5% per year with the interest paid at the end of each year:

EUR

Simple interest will usually be applied to savings accounts for payments within the year. Assume that on 12 February EUR 200 is deposited into a savings account with 6% interest. If you look at the account on 7 July nothing will have happened. The balance is still EUR 200. However, if you close the account on that day, you get back your original EUR 200 plus simple interest for the time you left the money in the bank.

Simple interest is usually calculated on a daily basis. In Germany, the banking year has 360 days and 12 months, meaning that every month has the same number of 30 days. Another regulation is that either the day of in-payment or the day of withdrawal can be counted as an interest day, but not both.

nI D i n= ⋅ ⋅

200 0.05 10I D i= ⋅ = ⋅ =


Between 12 February and 7 July we can therefore count 145 interest

days. For each interest day interest

is paid, i.e. for 200 EUR and 145 days: 4.83 EUR.

We will refer to simple interest again in Section 6.2, when we discuss savings.

6.1.2 Compound Interest

In Volume 1, Section 4.4.3 we learned about compound interest in connection with the exponential function. Compound interest is the concept of adding accumulated interest back to the principal, so that interest can also be earned on interest from that moment on. Declaring that interest is part of the principal is called compounding (i.e. interest is compounded). The interest may be compounded every year, but also in shorter (or even longer time periods). If compounded quarterly, the interest is added to the principal every quarter (= 3 months); monthly (one month); or even daily. We can also say that interest is capitalised, i.e. added to the balance.

Interest rates must be comparable in order to be useful, and in order to be comparable, the interest rate and the compounding frequency must be known. Because most people think of rates as a yearly percentage, many governments require financial institutions to disclose a comparable yearly interest rate on deposits or advances. Compound interest rates may be referred to as effective interest (also: Annual Percentage Rate, Effective Annual Rate, and other terms). When a fee is charged up front to obtain a loan, effective interest usually counts that cost as well as the compound interest when calculating the converted rate. These government requirements assist consumers in comparing the actual cost of borrowing.

0.060.0001667 0.01667%

360 360

i = = =

6 . 1 I n t e r e s t P a g e | 63

t

D

interestperiod

interest rate i

.....balance after 1 period:

balance after 2 periods:

1 (1 )B D D i D i= + ⋅ = ⋅ +

2 1 1 1 (1 )B B B i B i= + ⋅ = ⋅ +balance after n periods:

1 (1 )n nB B i−= ⋅ +

Figure 6-3: Compound interest

Suppose a principal (or deposit) of D (in euro) yields interest at a rate i % per period. After the first period the balance will be:

with the interest rate i

The interest rate may either be given as a percentage of the principal, i.e. i = 5%, or – as is more common in practice and for most of this chapter – as a decimal, i.e. i = 0.05.

After another period, capital and interest will be:

After n periods the balance is:

This is the formula for compound interest. The term implies that interest paid (earned) on a deposit is periodically added to the principal. The term

is called the interest factor or often the compound factor. For each interest period the capital will be multiplied by q.

1 (1 )B D i= ⋅ +

22 1 (1 ) (1 )B B i D i= ⋅ + = ⋅ +

(1 )nnB D i= ⋅ +

1q i= +


FUTURE VALUE

A principal D, compounded with rate i will increase within n interest periods to:

nnB D q= ⋅ with q = 1+i being the interest factor

is also called the future value of the principal D after n interest

periods.

Both formulas (that of simple interest and that of compound interest) contain four basic financial terms:

• D = principal, deposit, or present value

• i = interest rate per time period (for instance one year), or in many cases the interest factor q

• n = number of periods (e.g. years)

• nB = balance after n interest periods, or future value

These four basic variables will occur in some form in every model. Three of the four terms are usually given, and the fourth is then found.

t

interest rate i

...

D

interest factor q = 1 + i

today

?nB =

eoy 1 eoy 2 eoy n-1

eoy n

eoy = end of year

Figure 6-4: Future value of money

nB

6 . 1 I n t e r e s t P a g e | 65

We will begin with the most obvious case: Looking for the future value of a principal D deposited n interest periods ago, we can calculate the future value of the deposit using the known interest rate and number of periods. Positioning "today" at the point in time for which we want to know the balance (see Fig. 6-4) produces the standard compound interest formula:

nnB D q= ⋅

EXAMPLE

Anna deposited D = 1000 EUR in a saving account with an interest rate of 6%. What is the balance after 8 years?

8 8 88 (1 ) 1000 1.06 1593.85B D i D q= ⋅ + = ⋅ = ⋅ = EUR

As in all problems of financial mathematics, it is advisable to check whether the result is roughly comprehensible or not. This is not only a nice way of verifying whether the calculations are correct or not, but is also effective training for your daily work.

Cross check

Anna will get at least EUR 60 annually. That means that in 8 years she will earn at least EUR 480. Interest on interest will add some additional amount to the balance. Therefore the result is probably correct.

We all realise that 1 euro today is worth more than 1 euro to be received one or two years from now, assuming that interest can be earned. Calculating the present value of future cash flows allows us to place all cash flows on one level field, or reference point, such that comparisons can be made about present values.


t

eoy=end of yearinterest rate i = 0.06

...

D =?

todayeoy 1

8 2500 EURB =

eoy 2 eoy 7

eoy 8

Figure 6-5: Present value of money

PRESENT VALUE

Finding the present value is simply the reverse of compounding:

Note that the term 1 11 i q+ = is the reciprocal of the interest factor q. It is

often called the discount factor, denoted by 1 11q iv += = . Therefore,

finding the present value (or discounting) of a given future value consists of multiplying by the discount factor for every interest period.

EXAMPLE

How much does Anna have to deposit if she wants to have a balance of EUR after 8 years, assuming a constant interest rate of

6%?

( )88 88 18 1.068

2500 2500 0.943396(1 )

BD B v

i= = ⋅ = ⋅ = ⋅ =

+

88 2500 0.62741 1568.53D B v= ⋅ = ⋅ = EUR

1 1

(1 )n

n n nn nD B B B v

i q= ⋅ = ⋅ = ⋅

+

8 2500B =

6 . 1 I n t e r e s t P a g e | 67

Sometimes we are faced with a situation in which we know the future and present values as well as the number of periods involved. What is unknown, however, is the compound interest rate i, implicitly supplied in the described situation. Again, three of the basic terms are known ( ); the fourth, the interest rate i, is unknown and should be

calculated.

Solving the formula of compounding interest for i results in:

→ →

INTEREST RATE

The interest rate i, given the present and the future value as well as the number n of interest periods, is:

EXAMPLE

We assume that Anna has only EUR 1400 to deposit. She wants to know what the interest rate must to be in order to reach a balance of EUR 2500 after 8 years?

Hence, the interest rate must be 7.5168%. Now she has to find a bank which offers this rate of interest.

, , and nD B n

(1 )n nnB D i D q= ⋅ + = ⋅

1

1nn n n nn

B B Bq q i

D D D = → = + = =

1

1nnB

iD

= −

1nBnD

i = −

18

82500 2500

1 1 1.075168 1 0.0751681400 1400

i = − = − = − =


Cross check

The interest rate must definitely be greater than 6%, based on Anna’s deposit. 7.5% is a realistic rate.

At times we may need to know how long it takes for an amount deposited today to grow to a certain future value at a particular compound interest rate. In this case the time is being calculated, in particular, the number of periods an amount has to be compounded in order to reach a certain balance.

This time, the three basic terms given are: , and we are

looking for the number of interest periods n until the required balance is reached.

Now we have to solve the compound interest formula for n. Since n occurs in the exponent we have to apply the logarithm (see: Section 5.2, Page 43) in order to invert the term, such that we can isolate the unknown. The sub index n of the symbol is certainly not the variable

n but only a part of the balance name.

NUMBER OF INTEREST PERIODS

The number of interest periods n to reach a given future value is:

ln ln

lnnB D

nq

−=

It should be clear that instead of the natural logarithm "ln", any other logarithm could also have been applied.

, ,and nD B i

nB

with 1nnB D q q i= ⋅ = +

ln ln ln lnn n nn

B Bq n q B D

D D= → ⋅ = = −

ln lnln ln ln

lnn

nB D

n q B D nq

−⋅ = − → =

6 . 1 I n t e r e s t P a g e | 69

EXAMPLE

Anna was not successful in finding another bank which offers her the 7.5% interest rate she needed to reach a balance of EUR 2500 within 8 years. Therefore, at an interest rate of 6%, she has to wait longer than 8 years until her deposit of EUR 1400 grows to EUR 2500. How long will it take?

D = 1400; ; q = 1.06

ln ln ln 2500 ln14009.95

ln ln1.06nB D

nq

− −= = = years

That means Anna has to wait almost 10 years.

Cross check

Anna did not find a bank offering a higher interest rate. The time until the final amount is reached must definitely be longer than 8 years. Since she has to earn about EUR 160 in interest, “two additional years at the rate of 6% for EUR 1400” sounds plausible.

ATTENTION

Please note, that in all calculations the intermediate results shown in the formulas are always displayed with a limited number of decimals, although the maximal number of decimals on the calculator are always used for calculations and final results. Final result and calculations with fewer decimals may therefore be slightly different.

6.1.3 Effective Interest

Up to now, we have assumed that interest is paid annually, as this assumption makes it easiest to get a basic understanding of the time value of money. However, we are quite regularly confronted with compounding periods shorter than a year. What is the relationship between future value and interest rates in these situations?

2500nB =


To begin with, suppose that the interest is paid semi-annually, that is, twice a year. We have already discussed the situation where compounding periods are less than a year and the number of compounding operations increases as time intervals become shorter (see Section 5.1.4; Page 28).

It is now very important to have a clear understanding of the relationship between time intervals, the number of compounding operations and the applied interest rate.

• If nothing else is mentioned, the interest period is generally a year. Any other period must explicitly be stated!

• The interest rate stated is usually the nominal interest rate, implicitly assuming an interest period of one year.

• In case of shorter compounding periods, the nominal interest rate has to be adapted accordingly.

If you deposit D = 100 EUR in a savings account at a nominal 9% interest rate, the future value at the end of the first half a year would be:

EUR

At the end of a full year, the future value would be:

EUR

This amount has to be compared with EUR, in which interest

was paid only once a year. The difference of 1 1 0.20B B B∆ = − =ɶ EUR is

caused by interest earned in the second half of the year on the interest of 4.50 EUR that was earned at the end of the first half a year. This is the effect of the compounding within the year.

We will now generalise the above problem by compounding p times in a year – which we already did in Section 5.1.4 when discussing the exponential function. We therefore divide the year into p periods. If we assume a nominal interest rate i and interest paid periodically at a rate

1i ip pp p

i q= → = + , the balance after p periods (= one year) is:

12 2(1 ) 100 (1 0.045) 100 (1.045) 104.50iB D= ⋅ + = ⋅ + = ⋅ =

121 (1.045) 104.50 (1.045) 109.20B B= ⋅ = ⋅ =

1 109B =ɶ

1 (1 )ppB D i= ⋅ +

6 . 1 I n t e r e s t P a g e | 71

t

interest rate per period

...

D

B1

pi

today

eop=end of period

eop 1 eop 2

eop p

eop p-1

Figure 6-6: Interest compounded p times per year

The following Tab. 6.1 contains the future values of the deposit D = 100 EUR at the end of the first year, with different compounding periods, and continuous compounding at a nominal interest rate of 8% (please refer to Volume 1, Section 4.4.4):

Periods p per year

1 3 6 12 360 continuous

in EUR 108 108.22 108.27 108.30 108.33 108.33

Table 6.1: Future values

The shorter the compounding period, the more periods there are and the higher the future value; the maximum possible value can be reached with continuous compounding − which does not differ from daily compounding inside the first two decimals.

A consumer who wants to take out a loan may be faced with several offers from competing financial institutions. It is therefore of considerable importance to compare the various offers, especially if they are based on different assumptions. The effective interest rate is often used in making such comparisons.

For instance, comparing the effective annual interest rates applied to Table 6.1, the interest paid (or earned) is:

1B


Periods p per year


Effectively paid interest

8% 8.22% 8.27% 8.3% 8.33% 8.33%

Table 6.2: Effective interest

It is clear that the yearly paid interest is higher when interest is compounded more often within the year.

If we want to compare investments that have different compounding periods, we need to state their interest using some common or standardised form. This leads us to distinguish between nominal interest rate and effective annual interest rate .

In banking, and even more generally in the lending business, the concept of effective interest rates is becoming more and more important.

EFFECTIVE INTEREST RATE

The effective interest rate is always the interest rate per year taking

into account the total cost factors.

Basically, there are three factors that can create a difference between nominal rate and the effective rate, though it should be obvious that in credit the effective interest rate is always higher than the nominal rate:

1. The compounding period may be less than a year, causing additional interest on interest.

2. Bank charges for services are not covered by the interest rate. These fees are usually collected once, generally at the beginning of the contract; sometimes just as a fixed cost, sometimes also depending on the amount (of a loan, for example). In some cases, fees might even be collected several times during the contract. Fees are costs for the client and therefore increase the total cost of the money borrowed.

effi

effi

6 . 1 I n t e r e s t P a g e | 73

3. The nominal interest rate changes during the contract time (floating interest). In this case, the effective interest rate is the average interest rate for the entire contract time when compounding yearly. However, “average” does not mean that the average is calculated as the arithmetic mean of the interest rates!

We often see a combination of all three approaches.

The effective interest rate is always calculated as a yearly rate on the basis of all costs charged to the client. In Germany, it is set in the law that the effective interest rate must be explicitly stated on every contract involving interest rates. As can be seen, calculating the effective interest rate is a very important topic in banking and finance.

COMPOUNDING PERIODICALLY

If the interest is compounded periodically for p periods during a year at a nominal rate i, the interest earned on interest influences the effective rate. Shorter compounding interest periods increase the effective interest.

D

t

B1

today

ip p

i =

eop 1 eop 2 eop p-1

eop p

interest rate per period

interest rate per year (nominally) i

eop = end of period

Figure 6-7: Compounding periodically

Let us assume that a deposit is compounded p times per year. After 1 year the balance will be:


( )1 1p

ipB D= ⋅ +

Because this has to be the same amount if compounding yearly with the effective interest rate, the following equation arises:

D can be cancelled out, which means that the effective rate is independent of the amount deposited.

EFFECTIVE INTEREST RATE : PERIODICALLY COMPOUNDING

The effective interest rate is independent of the amount D. For a given nominal interest rate i, it increases with the number of periods p:

For continuous compounding, the effective interest rate is:

The same argumentation holds if we consider more than just one year – which is commonly the case.

Let us assume again that a deposit is compounded p times per year. After n years, the balance will be (see Volume 1, Section 2.2.1, or Volume 1, Section 4.4.2):

This balance must also be the result when we compound the principal D effectively over n years, because the effective interest rate is always the correct rate (coving all adjustments) for exactly one year:

The right-hand sides of the last two equations must also be equal:

( )(1 ) 1p

ieff p

D i D⋅ + = ⋅ +

( )1 1p

ieff pi = + −

1ieffi e= −

( )(1 ) (1 )n

p n pi in p p

B D D ⋅= ⋅ + = ⋅ +

(1 )n nn eff effB D q D i= ⋅ = ⋅ +

6 . 1 I n t e r e s t P a g e | 75

( ) ( )1 1n p ni

n effpB D D i⋅

= ⋅ + = ⋅ +

Cancelling the deposit D (the effective interest rate should not depend on the deposit!) and removing the n-th root from both sides will result in:

The same result is found as before:

EXAMPLES

1. What is the effective yearly rate corresponding to a nominal

annual interest rate of i = 10% with interest compounded:

a) each quarter b) each month?

a) p = 4 → : ( )40.141 1 0.1038effi = + − =

effective interest rate is 10.38%

b) p = 12 → : ( )120.1121 1 0.1047effi = + − =

effective interest rate is 10.47%

Cross check

In both cases, the effective interest rate must be higher than the nominal 10% − which is the case. Also, and the rate in case b) must be higher than in case a), because the compounding periods are shorter. This is also true.

(1 ) (1 ) (1 ) 1n p n pi inneff effp pi i⋅+ = + → + = +

(1 ) 1pieff pi = + −

effi


2. The deposit of D = 200 EUR is compounded periodically by p=1, 3, 6, 12, 360 periods and continuously.

a) Calculate the balance after 3 years at the nominal interest rate of 10%.

b) Calculate the effective interest rates of the different compounding periods: p = 3, 6, 12, 360, continuous

Periods p per year


a) in EUR 266.20 268.65 269.31 269.64 269.96 269.97

b) effi 0.1 0.1034 0.1043 0.1047 0.1052 0.1052

Table 6.3: Effective interest rates

3. The monthly interest for micro loans with 6 months maturity is 2%. What are the nominal and the effective interest rates for this loan?

The nominal interest rate is:

Despite a maturity of only 6 months, the effective interest rate is always calculated on the basis of a complete year! That means that the effective interest rate is:

Cross check

The nominal interest rate is the yearly rate, i.e. the monthly rate times 12. The effective interest rate is always greater than the nominal rate: 26.8% is convincing.

Sometimes it is interesting to calculate the interest rate for a given period so that a certain effective rate can be reached. Simply invert the question and, consequently, the calculation: the result is the effective interest rate.

12 12 2% 24%mi⋅ = ⋅ =

12 12(1 ) 1 1.02 1 0.2682 26.82%eff mi i= + − = − = =

6 . 1 I n t e r e s t P a g e | 77

What should be the monthly (quarterly, or weekly) interest rate, so that the corresponding compounding leads to the intended target value? In order to answer this question we use the above formula:

and solve it for pi :

→

EXAMPLE

The effective rate of is the result. What is the equivalent

interest rate if you compound

a) quarterly? → , i.e. 1.943%

b) monthly? → , i.e. 0.6434%

c) daily? → ,

i.e. 0.02134%

(1 ) 1 (1 ) 1p pieff ppi i= + − = + −

(1 ) 1pp eff effi i q+ = + = 1 1p p

p eff p effi q i q+ = → = −

8%effi =

44 1.08 1 0.01943i = − =

1212 1.08 1 0.006434i = − =

1360360

360 1.08 1 (1.08) 1 0.0002134i = − = − =


EXERCISE 6.1: INTEREST

1. Assume the nominal interest rate is 8%. How much simple interest will be paid on:

a) EUR 200 for 5 months? b) EUR 600 for 127 days?

c) EUR 800 for the time between 18 March (day of deposit) and 10 October (day of closing the account)?

2. An investment of EUR 1500 is compounded 10 times at an interest rate of 4.5%. What will the final balance be?

3. EUR 200 is deposited into a savings account at 5% interest. What will the value of this money be after 6 years?

4. EUR 500 is invested in a savings account with 1% interest compounded monthly. What will the balance of this account be after 15 months?

5. The balance of an account is EUR 785.34. How much must have been invested 3 years ago, if the interest rate was a constant 4%?

6. How much should you invest today if you need EUR 2000 in 5 years at a fixed interest rate of 7%?

7. Your account (0.75% interest compounded monthly) has a balance of EUR 3211.56. What was the balance 15 months ago?

8. You have invested EUR 1250.

a) Calculate the interest rate, so that the principal doubles within 10 years, without additional deposits.

b) Assuming monthly compounding, calculate the monthly interest rate.

6 . 1 I n t e r e s t P a g e | 79

9. How long does it take to double your investment of EUR 1250 at an interest rate of 6%? (Without additional clarification the interest rate is understood to be a yearly compounding rate. Thus the number of interest periods, and therefore the answer, should be in years.)

10. The balance of your saving account is EUR 278.66. How long do you have to wait until the balance is EUR 400, assuming a monthly compounding interest rate of 0.8%?

11. Calculate the effective interest rate for:

a) a monthly compounding rate of 1.5%

b) a quarterly compounding rate 0f 4.5%

c) a bi-monthly rate of 4%

12. The nominal yearly interest rate is 9%. Calculate the effective interest rate for:

a) quarterly compounding

b) monthly compounding

c) daily compounding, assuming that the banking year has 360 days

13. The effective interest rate is 8.5%. Calculate the interest rates for the following compounding periods which lead to this given effective rate:

a) quarters b) months c) days


ANSWERS 6.1: INTEREST

1. a) EUR b)

c) 202 interest days → EUR

2. EUR

3. 6 268.02B = EUR

4. 15 580.48mB = EUR

5. I = 698.16 EUR

6. I = 1425.97 EUR

7. B = 2871.05 EUR

8. a) i = 7.177% b)

9. n = 11.9 years

10. n = 45.36 months

11. a) b) c)

12. a) b) c)

13. a) b) c) 0.0227%effi =

5 6.67mI = 127 16.93dI =

202 35.91dI =

10 2329.45B =

0.5793%mi =

19.56%effi = 19.25%effi = 26.53%effi =

9.31%effi = 9.38%effi = 9.42%effi =

2.06%qi = 0.682%mi =

6 . 1 . 4 P r o g r e s s T e s t " I n t e r e s t " P a g e | 81

6.1.4 Progress Test “Interest”

You should allot yourself some time for concentrated work on this test. Try to solve as many problems as you can. Do not use the reader to look for the solution. The aim of the test is to get feedback on how much you know or have learned up to now.

At the end of the chapter you will find the solutions to the problems. Each solution is reduced to the final answer; it may even be only a number, a symbol, a table or a graph. You should check your answers. If they are correct, you may continue and start the next chapter. In any other case (you got the wrong answer or no answer at all) you should repeat the corresponding units in order to close the gap.

1. Anna opens a savings account at 8% interest and deposits EUR 300 on 25 May. She closes the account on 6 November. How much will she receive, provided simple interest is paid?

2. Anna invests EUR 500 at a nominal interest rate of 10%. What will the balance be after 6 years, if interest is compounded:

a) annually? b) quarterly? c) monthly?

3. What will the effective interest rates be for Problem 2?

4. Anna would like to deposit a certain amount so that she has EUR 800 in her account after three years at a rate of 0.57%, compounded monthly. How much does she have to deposit?

5. How long must she wait under the conditions of Problem 4, if she has only EUR 550 for deposit?

6. Procredit agrees to increase the interest rate. What must the new monthly interest rate be so that Anna receives EUR 800, although she deposited only EUR 550 three years ago?


7. The effective interest rate is 14.5% for a ProCredit loan account. What is the equivalent interest rate compounded:

a) quarterly? b) bi-monthly? c) monthly?

6 . 2 D e p o s i t s

6.2 Deposits

Prerequisites: The concept of interest and calculations of cashas a function of time are essential for the calculation of deposits.

Learning Targets: The basic rules of discussed. We value of single investments adiscussion to leads to another important tool mathematics: the application of geometric series.


6.1 Interest

Simple Interest

Compound Interest

Effective Interest

6.2 Deposits

Saving Rates

P a g e | 83

The concept of interest and calculations of cash flow function of time are essential for the calculation

The basic rules of handling deposits will be discussed. We will begin by calculating the futire value of single investments add then extend the discussion to multiple deposits. Saving at fixed rates leads to another important tool in financial mathematics: the application of geometric series.


6.2 Deposits

Savings

Saving Rates

6.3 Loans

Repayments

Instalments

Effective Interest


The two most basic operations of our banks are deposits and loans. Both operations rely heavily on the concept of the “time value of money”, i.e. interest. In the case of deposits, the client provides the money and the bank owes it and thus pays interest. In the case of loans it is just the opposite, the bank provides (disburses) the money and the client owes it and pays interest for this service. In principle there is no difference when analysing the processes. The procedures of dealing with interest is identical in both situations – the distinction is only in perspective: in the case of deposits, the client is the owner and the bank owes money. In the case of loans, it is just opposite.

In this section we will cover the basic operations of saving. Many aspects were already discussed in the last section. In order to avoid too much redundancies, we refer often to the results of the former analyses.

6.2.1 Savings

Savings is a deposit in a specially designed account. Money is deposited into the account, and beginning the next day, interest is calculated. The client knows the nominal rate. In most cases this is equal to the effective interest rate, because in most countries the interest is paid (capitalised = added to the balance) on a yearly basis at the end of the calendar year, and there are usually no fees on saving accounts.

SINGLE PAYMENTS

For deposits within a year, simple interest is paid (see Section 6.1.1; Page 61) only if the account is closed. In this case, the balance B is paid out using the deposit D plus the interest earned for the days between in-payment (not included) and the day of out-payment (included). The difference represents the number of interest days d.

6 . 2 D e p o s i t s P a g e | 85

BALANCE AT SIMPLE INTEREST

Simple interest is paid for the deposit D at the nominal interest rate i for d interest days:

( )360 3601i i

dB D D d D d= + ⋅ ⋅ = ⋅ + ⋅

t

interest rate i

D

Bd

interest factor q = 1 + i

today

d interest days

Figure 6-8: Savings account

EXAMPLE

Anna opens an account with a 5% interest rate by depositing EUR 400 on 25 May, and then close it on 12 October. What will the balance be?

On the basis that each month has 30 days, day of in-payment excluded, day of out-payment included, we obtain 5 days in May + (4 months =) 120 days + 12 days in October = 137 interest days.

( )0.05137 360

400 1 137 407.61B = ⋅ + ⋅ = EUR

Cross check

5% for the whole year would be EUR 20. Four months is a third of the year, i.e. roughly EUR 7. Therefore the result is probably correct.

If the account exists beyond the end of the year, the interest earned during the old year is capitalised on 31 December and then added to the account. From the next day on, interest is paid also for the added interest.


The capital at the end of the old year with od interest days is:

In the new year, interest is paid on for interest days:

EXAMPLE

Anna opened an account with a 5% interest rate by depositing EUR 400 on 25 May 2007, and then close it on 18 April 2008. What will the balance be?

Number of interest days in 2007: 5 days in May + (7 months =) 210 days in 2007 = 215 days

Number of days in 2008: (3 months =) 90 days + 18 days in April = 108 days

EUR

Cross check

As in the previous example, interest for one complete year will be EUR 20. More than 7 months, roughly three qaurters, i.e. EUR 15 interest, plus interest on interest. EUR 18 is probably correct.

SUCCESSIVE PAYMENTS

Suppose that three successive deposits are made for the amounts

1 400D = EUR at the beginning of the first year, then 2 700D = EUR one

year later, and 3 300D = EUR another year later. How much will be in

the account at the end of year 3, given an interest rate of 6% per year?

According to our notation, the three deposits are positioned in Fig. 6-9 as arrows from above. We now have multiple payments plus balancing.

360(1 )od i

oB D ⋅= ⋅ +

oB nd

0360 360 360

(1 ) (1 ) (1 )n nd i d i d in oB B D⋅ ⋅ ⋅= ⋅ + = ⋅ + ⋅ +

215 0.05 108 0.05360 360400 (1 ) (1 ) 418.12B ⋅ ⋅= ⋅ + ⋅ + =

6 . 2 D e p o s i t s P a g e | 87

t

interest rate per year i

today

1 400D = 2 700D = 3 300D =

3B

Figure 6-9: Successive payments

Since the cash flow changes with time, we can only add, subtract, compare, and balance cash flows when all the payments are adjusted to one point in time. This date is usually marked extra as the reference point, or often simply by the word “today”.

If we want to know the account balance after 3 years, we must position the reference point at the end of the third year. The only present value is the balance . The three deposits are past values.

BASIC RULES FOR BALANCING

There are two basic rules for calculations in finance, which always have to be taken into consideration:

1. All payments have to be adjusted to one point in time. After this point in time is fixed, all past payments have to be compounded, all future payments have to be discounted, and today’s payments are present values.

2. The sum of all compounded (or discounted) in-payments must be equal to the sum of all compounded (discounted) out-payments.

Payments in the above sense are not necessarily physical cash flows. For instance, if there are many deposits to an account and we ask for the

3B


balance at a certain time, then we assume a withdrawal will comply with Rule 2 above (balancing being an out-payment).

In the above example, we refer all payments to “today” (Rule 1):

Sum of deposits: (present value)

Sum of payments:

compounded one year → with being the

compound factor

compounded two years →

compounded three years →

Rule 2 states that the sum of all in-payments and the sum of all out-payments should be the same:

Thus, with q =1.06 the balance after 3 years is:

EUR

If we position our reference point differently (for instance in Fig. 6-10 at the first deposit) then the calculation will be different; however, the result will be the same.

t

interest rate per year i

today

1 400D = 2 700D = 3 300D =

3B

Figure 6-10: Positioning "today"

3B

3D 3D q⋅ 1q i= +

2D 22D q⋅

1D 31D q⋅

2 33 3 2 1B D q D q D q= ⋅ + ⋅ + ⋅

3 318 786.52 476.41 1580.93B = + + =

6 . 2 D e p o s i t s P a g e | 89

If we balance all payments to the point of the first deposit, we obtain:

with 1 11q iv += = being the discount factor

Multiplying this equation by and bearing in mind that , we can

prove that the balance is the same as before.

6.2.2 Regular Equal payments

If all payments are equal ( ) then the value of all

deposits can be calculated by applying the geometric series. Since equal rates of payment and deposits occur quite often, we will discuss this in more detail.

EXCURSION

The geometric series is often applied in financial mathematics. It is the sum of terms, starting with the number 1 and always increasing by a constant factor q. The geometric series with the constant factor q and n terms is:

Note that there are n term although the last exponent is because we

start with . This sum occurs very often, with q being the interest factor. There is a very convenient formula that avoids calculating the sum:

Arriving at this formula is not difficult but is, however, beyond the scope of this course.

An annuity is a sequence of equal payments made at set intervals over some time span (although the word “annuity” suggests annual payments,

2 31 2 3 3D D v D v B v+ ⋅ + ⋅ = ⋅

3q 1qv =

3B

1 2 3D D D a= = =

2 3 11 ... nq q q q −+ + + + +

1n−01 q=

2 3 1 1 11 ...

1 1

n nn q q

q q q qq q

− − −+ + + + + = =− −


this is not necessarily the case). Let us assume there are n such in-payments at the beginning of each year. What will the balance be at the end of the nth year?

t

Bn

a

today ....

a a a

interest rate per year i interest factor per year q =1+ i

Figure 6-11: Regular equal payments

Assuming n regular deposits of value a at the start of each year, we are interested in the balance at the end of year n. The interest rate i is fixed.

Balancing all in-payments requires that we set a reference point in time. Since we are interested in the final balance, it is quite natural to select the end of year n as the reference point, and adjust all payments to this point in time. Because all payments are past payments, they have to be compounded.

• The last payment is one year old; it has to be compounded once: → a q⋅

• The preceding payment is two years old; it has to be compounded

twice: →

• The preceding payment is three years old; it has to be

compounded three times: →

• … and so forth …

• The first payment is n years old; it has to be compounded n times:

→

Now we have adjusted all in-payments and their sum will be the balance:

2a q⋅

3a q⋅

na q⋅

6 . 2 D e p o s i t s P a g e | 91

{ }2 3 2 1... 1 ...n nnB a q a q a q a q a q q q q−= ⋅ + ⋅ + ⋅ + + ⋅ = ⋅ ⋅ + + + +

The common factor can be excluded. The sum in the brackets is the geometric series resulting in:

{ }2 2 1... 1 ...n nnB a q a q a q a q q q q−= ⋅ + ⋅ + + ⋅ = ⋅ ⋅ + + + +

1 1

1

n n

nq q

B a q a qq i

− −= ⋅ ⋅ = ⋅ ⋅−

This formula is one of the most used calculations in financial mathematics. It occurs regularly whenever constant payments at regular intervals arise: savings, loans, insurance, pensions, etc.

BALANCE AFTER N INSTALMENTS

Assuming regular payments of a per period and an interest rate i (→ compounding factor ), the balance after n periods is:

EXAMPLES

1. Anna saves EUR 500 every year (in-payment at the beginning of the year) in an account with 6% interest. What is the balance after 8 years?

a = 500; n = 8; i = 0.06 → q = 1.06

Balance: EUR

Cross check

Sum of all in-payments: EUR

a q⋅

1q i= +

1 1

1

n n

nq q

B a q a qq i

− −= ⋅ ⋅ = ⋅ ⋅−

8

81.06 1

500 1.06 5245.660.06

B−= ⋅ ⋅ =

8 500 4000⋅ =


6% per year for in the average EUR 2000:

EUR

Taking into account the compounding effect, the result is plau-sible.

2. Anna saves EUR 10 monthly in your child’s account. How much will be in the account when your child turns 18, assuming a constant interest rate of 0.5% for the duration of the account?

a = 5; n = 18 times 12 months = 216 months; i = 0.005

→ q = 1.005: ?nB =

216

18 2161.005 1

10 1.005 3881.280.005y mB B

−= = ⋅ ⋅ = EUR

Cross check

Yearly investment: EUR 120 times 18 years:

EUR 2160 in deposits

Average balance of the account: EUR 1080

Interest per year: at least 6% per year → interest at least

1080 0.06 18 1166⋅ ⋅ = EUR

The balance must be in the range of 3326 EUR, which is the case.

The formula for regular successive payments once

more shows the four basic financial items:

• the future value: the balance

• the present value: the payment a

• the time: the number of periods n

0.06 2000 8 960⋅ ⋅ =

1n

nq

B a qi

−= ⋅ ⋅

nB

6 . 2 D e p o s i t s P a g e | 93

• the interest rate: the interest rate i, or the compound factor q, respectively

Again, three must be given in order to calculate the fourth.

The balance is already isolated in the above formula.

If we want to know how much we should pay periodically in order to obtain a certain balance given the time n and the interest rate i , we have to solve the formula for a:

EXAMPLE

How much does Anna have to save monthly if the monthly compounding interest rate is 0.5% and she wants to reach EUR 2000 within 18 months?

EUR

Cross check

Saving EUR 2000 within 18 months requires monthly in-payments of approximately EUR 100, resulting in EUR 1800 plus interest. The above result makes sense.

Assume Anna cannot afford the EUR 105.93 monthly but instead EUR 85. How long does she have to deposit that regular monthly amount in order to reach the EUR 2000, provided the interest rate does not change?

Now the present and the future values are given, as well as the interest rate. We are looking for the time, and have to solve the formula for n:

( 1)nn

B ia

q q

⋅=⋅ −

2000; 18; 0.005; 1.005 : ?nB n i q a= = = = =

( )18

2000 0.005105.93

1.005 1.005 1a

⋅→ = =⋅ −


1n

nq

B a qi

−= ⋅ ⋅ → 1 nnB iq

a q

⋅+ =

⋅

We need to solve this equation for n, which is the exponent of the compound factor. We have to apply the inverse operation of the exponential operation, and that is the logarithmic operation. We can choose any logarithm, for instance:

ATTENTION

It is not possible to simplify the expression in the numerator of the above fraction. A common mistake is to solve the logarithm in the numerator as follows:

It is easy to realise that the above calculation must be wrong, because and the remaining two expressions cannot be equal!

EXAMPLE

We will try to find a solution to Anna's problem:

months

ln 1

ln ln 1 ln ln 1ln

n

n n n

B i

a qB i B iq n q n

a q a q q

⋅ + ⋅ ⋅ ⋅ = + → = + → = ⋅ ⋅

ln 1 ln ln1n nB i B i

a q a q

⋅ ⋅+ = + ⋅ ⋅

ln1 0=

85; 2000; 0.005; 1.005 : ?na B i q n= = = = =

2000 0.005ln 1

0.11070285 1.00522.2

ln1.005 0.0049875n

⋅ + ⋅ = = =

6 . 2 D e p o s i t s P a g e | 95

Cross check

Every month Anna deposits about 20% less; i.e. she has to wait about 20% longer. 20% of 18 months is about 4 mothns: The 22 months seem to be correct.

Now we would expect to discuss the fourth problem, calculating the interest rate, provided the present and future values as well as the time. In this case, a polynomial equation of degree n + 1 has to be solved. This is a very complex problem, and can only be accomplished by means of an advanced approximation technique. Therefore, the discussion of this problem must be omitted here.


EXERCISE 6.2: DEPOSITS

1. How many interest days have to be taken into account if the day of deposit is 17 April 2007 and the day of repayment is 5 September 2007?

2. Anna deposits EUR 370 into a savings account at 5% nominal interest on 7 June, and then withdraws it on 6 December. How much will she receive?

3. An investment of EUR 500 made on 4 Februaray will become EUR 530.56 within 188 interest days, provided simple interest is paid. What is the nominal interest rate?

4. Anna makes four payments: at the beginning of the 1st year: EUR 300; the 2nd year EUR 600; the 4th year EUR 500, and the 6th year EUR 800. How much will be in the account at the end of the 8th year, given an interest rate of 5% per year?

5. Together with the payments from Problem 4, Anna withdraws EUR 400 at the beginning of year 5 and EUR 300 at the beginning of year 7. How much will now be in the account at the end of year 8, given an interest rate of 5% per year?

6. Use the payments from the following figure, given an interest rate of 6%:

ttoday

1 400D =4 480D =

0D

2 800W = 3 350W =5 660W =

eoy5eoy1

eoy2 eoy3eoy4

eoy = end of yearinterest rate per year i=0.06

6 . 2 D e p o s i t s P a g e | 97

How much should be deposited "today" in order to cover all payments?

7. Use the formula for the geometric series to calculate the sums:

a)

b) 2 3 4 5 6 7 81 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3s= + + + + + + + +

c)

8. How much does Anna save within 6 years if she deposits EUR 25 every month at an interest rate of 0.6%, compounded monthly?

9. Assume an effective interest rate of 8.5%. You save EUR 50 monthly. What will the balance of your account be after 5 years, if the interest is compounded monthly?

10. How many months does Anna have to invest EUR 50 in order to save EUR 2500 at a monthly compounding interest rate of 0.8%?

11. How much does Anna have to save monthly at a monthly compounding interest rate of 0.8%, if she needs EUR 2500 after three years?

12. How much do you have to save quarterly if you need EUR 5000 within 2 years? The effective interest rate is 8% and interest is compounded quarterly.

( ) ( ) ( ) ( ) ( )2 3 4 5 61 1 1 1 1 12 2 2 2 2 2

1s = + + + + + +

2 3 4 51 1.1 1.1 1.1 1.1 1.1s = + + + + +


ANSWERS 6.2: DEPOSITS

1. Total: 138 interest days

2. Balance = 379.20 EUR

3. Nominal interest rate:

4. Balance: EUR

5. Balance: EUR

6. Deposit today: EUR

7. a) s = 1.9844

b) s = 1.4285

c) s = 7.7156

8. Balance: EUR

9. Balance: EUR

10. Time: n = 41.94 months

11. Monthly payment: EUR

12. Quarterly payment: EUR

0.11704 11.7%i = =

2851.74B =

2034.79B =

0 741.49D =

6 2256.58B =

5 3716.87B =

59.72mr =

572.61qr =

6 . 2 . 3 P r o g r e s s T e s t “ D e p o s i t s ” P a g e | 99

6.2.3 Progress Test “Deposits”

You should allot yourself some time for concentrated work on this test. Try to solve as many problems as you can. Do not use the reader to look for the solution. The aim of the test is to get a feedback on how much you know or have learned up to now.

At the end of the chapter you will find the solutions to the problems. Each solution is reduced to the final answer; it may even be only a number, a symbol, a table or a graph. You should check your answers. If they are correct, you may continue and start the next chapter. In any other case (you got the wrong answer or no answer at all) you should repeat the corresponding units in order to close the gap.

1. What will the balance of Anna's savings account be at 6% nominal interest at the end of year 8 if she

• deposits EUR 550 at the beginning of year 1,


• withdraws EUR 400 at the beginning of year 3,


• withdraws EUR 300 at the beginning of year 7?

2. Anna expects a tax refund of EUR 350 on 1 September, which she deposits in her account at 0.75% compounded interest. She wants to buy a notebook for EUR 980 at the end of the year. How much must she deposit on 1 January?

3. Beginning 1 January, Anna wants to save EUR 80 monthly at 0.75% compounded interest. What will the balance be at the end of the year?


4. How much would Anna have to save monthly until the end of the year with 0.75% compounded interest if the balance of her account at the beginning of the year is EUR 352.12, and she does not want to spend the tax refund (of problem 2) for purchasing the notebook at the end of the year?

5. Anna opens a new account with a 9% effective interest rate. How much must she save monthly so that she can buy the notebook at the end of the year?

6. Anna cannot afford the monthly amount she would have to pay according to Problem 5. She can save only EUR 50 per month. How long must she wait for the notebook?

6 . 3 L o a n s

6.3 Loans

Prerequisites: The concept of interest and calculations of cashas a function of time are also the bascalculations.

Learning Targets: The basic rules of We will begin with at once, and repayments. The most common way of repaying loans is with fixed calculation of instalments.


6.1 Interest

Simple Interest

Compound Interest

Effective Interest

6.2 Deposits

Saving Rates

P a g e | 101

The concept of interest and calculations of cash flow as a function of time are also the basis of loan calculations.

The basic rules of handling loans will be discussed. begin with the repayment of the priciple all and then extend the discussion to several

repayments. The most common way of repaying loans is with fixed amounts, which leads to the calculation of instalments.


6.2 Deposits

Savings

Saving Rates

6.3 Loans

Repayments

Instalments

Effective Interest


The fundamental result of the last chapter was that all cash flow has to be adjusted to a single point in time. Therefore, all past payments have to be compounded, and all future payments have to be discounted. These basic rules do not differentiate between in-payments (deposits) and out-payments (withdrawals or credits). Therefore it may not be surprising that the differences between deposits and credits are only minor. The general ideas are identical.

A loan implies a disbursement (out-payment) at the beginning of an agreement, followed by in-payments until the loan is repaid. Usually, a loan is not repaid all at once, but in the form of instalments. Depending on the contract, they might be identical for each period (year or month) or differ periodically.

6.3.1 Repayments

Each repayment always consists of two parts:

• The principal repayment, i.e. the money owed to the lender.

• The interest, which is the cost of lending the money.

We will begin the discussion with the easiest case: the total principal is repaid all at once as a single amount.

SINGLE REPAYMENT

The loan is disbursed at the beginning of the time interval. We assume that the principal will be repaid after n periods, given the compound interest rate i per period (see: Fig. 6-12).

6 . 3 L o a n s P a g e | 103

t

interest rate per period i

P

eop 1

R

....today

eop = end of period

eop 2 eop n-1 eop n

Figure 6-12: Single repayment

We set the reference point "today" at the moment of repayment. R is the only in-payment and is a present value in this situation. The principle P is the only out-payment, and is a past value, and therefore has to be compounded. n periods have passed since the loan was disbursed. Therefore:

If we compare this formula with the formula for the future value of some deposit D (see: Section 6.1.2; Page 63), we see that they are practically identical. The difference is simply the direction of the payments.

Again, we have the four basic financial entries: present value (P), future value (R), interest rate (i), and time (number of periods n). We can only calculate one of the items if the other three are given.

In practice, single repayments of loans are exceptional. Usually a loan will be repaid in more than one – often many − repayments.

SEVERAL REPAYMENTS

Let us assume that the principal is repaid in three instalments: the first after two periods, the second after three periods, and the last one

after four periods. After this last repayment, the principal should be

completely repaid (see: Fig. 6-13). The interest rate i is compounded per period.

(1 )n nR P i P q= ⋅ + = ⋅

2R 3R

4R


t

interest rate per period i

P

4R

today

eop = end o period

3R2R

Figure 6-13: Several repayments

We set the reference point at the time of maturity and balance the payments. Except for , all payments are past payments, which have to

be compounded. The loan is four periods old and has to be compounded four times. The balance of all in-payments and all out-payments equals:

Given the principal and, for instance, the first two repayments as well as the interest rate, we could calculate the last repayment . However, it is

not possible to calculate all three repayments knowing only the principal and the interest rate. There are too many unknowns to solve the equation.

REPAYMENT PLAN

Each repayment consists in part of interest on the outstanding principal, and of some principal repayment. The repayments are called instalments. The interest portion is largest in the first instalment because interest has to be paid on the whole disbursed amount. If the interest rate is fixed, the interest portion of the instalment will decrease in the following periods. At the end of the contract, the instalment consists mainly of principal repayments and only a small amount of interest, because the principal amount owed decreases as payments are made.

Instalment payments are prevalent in mortgage loans, auto loans, consumer loans, and certain business loans. Amortisation of a loan

4R

2 44 3 2R R q R q P q+ ⋅ + ⋅ = ⋅

4R

6 . 3 L o a n s P a g e | 105

involves determining the periodic payment necessary to reduce the principal amount to zero at maturity, while providing interest payments on the unpaid principal.

During negotiations about a loan, the bank usually provides a complete repayment plan, which displays all the relevant information about the loan and its repayment: the period, the outstanding amount, the interest, the principal repayment, and – the sum of interest and principal repayment – the instalment (the latter four entries all in the corresponding period).

A person borrows P = 50,000 EUR (= principal) at the beginning of a year and is supposed to pay it off in 5 instalments at the end of each following year, with interest at i = 10% compounded annually.

Most regular clients want equal payments over a specified number of periods, even though the repayment plan would look essentially the same if the instalments were not constant. In the above case, the yearly instalment payment would be a = 13189.87 EUR.

The repayment plan is shown in Tab. 6.4. In fact, the principal will be repaid in five equal instalments.

p OA(p) Int(p) PR(p) Inst(p) Year EUR EUR EUR EUR

1 50000.00 5000.00 8189.87 13189.87

2 41810.13 4181.01 9008.86 13189.87

3 32801.27 3280.13 9909.74 13189.87

4 22891.53 2289.15 10900.72 13189.87

5 11990.81 1199.08 11990.73 13189.81

Sum 159493.74 15.949.37 49999.92 65949.29

Table 6.4: Repayment plan1

1 p=periods; OA(p)=Outstanding Amount in period p; Int(p)=Interest to be paid in period p; PR(p)=Principle repayment in period p; Inst(p)=Instalment to be paid in period p


2000

4000

6000

8000

10000

12000

14000

year 1 year 2 year 3 year 4 year 5

InterestPrincipal repayment

50004181

32802289

1199

8189 9009 9910 10901 11991

Pay

men

t

Principal Repayment vs. Interest

Figure 6-14: Graph of the repayment plan

In Fig. 6-14, the client’s yearly payments are represented by columns of equal height, because the instalments are constant. The portion of interest in that amount is shaded and the principal repayment is the white part. Because the interest rate is fixed, the interest portion decreases with every principal repayment. Correspondingly, the interest increases over time.

The example raises the question of how to calculate instalments, which are also called annuities when we mean yearly instalments.

6.3.2 Instalments

The principal amount P is now supposed to be repaid in n equal instalments at the end of each year, with an interest rate i compounded annually.

6 . 3 L o a n s P a g e | 107

t

interest factor per year q=1+i

eoy 1 .........

P

Bn= 0

today

eoy 2 eoy neoy n-1

eoy = end of year

a

interest

pricipalrepayment

a

interest

pricipalrepayment

a

interest

pricipalrepayment

a

interest

pricipalrepayment

.........

Figure 6-15: Instalments

Fig. 6-15 represents the disbursed principal as the only out-payment (arrow downwards) P. The constant yearly payments a are in-payments, represented by arrows from above. The interest payment each year is i on the outstanding balance; it is the dotted decreasing part of the instalment. The complementary part of the instalment (bold) is the principal repayment in the corresponding year, which is subtracted from the outstanding balance left over from the previous year. At the end (maturity) the balance should be zero.

At the beginning of the first year, the outstanding amount is the principal. However, at the end of that year, the first instalment a is paid. This payment reduces the outstanding amount (balance ) at the end of year 1

(= start of year 2) to:

Note that we simply balance all compounded in and out-payments.

Similarly, at the end of year 2 (= start of year 3), we balance:

Thus, at year j the balance will be:

1B

1B P q a= ⋅ −

22 1B B q a P q a q a= ⋅ − = ⋅ − ⋅ −

11 ...j j

j jB B q a P q a q a q a−−= ⋅ − = ⋅ − ⋅ − − ⋅ −


t

interest factor per year q=1+i

eoy 1 ...

P

a

int

rep

Bn=0

today

1B P q a= ⋅ −2

2 1B B q a P q a q a= ⋅ − = ⋅ − ⋅ −1

1 ...j jj jB B q a P q a q a q a−

−= ⋅ − = ⋅ − ⋅ − − ⋅ −

a

int

rep

aint

rep

aint

rep

... eoy 2 eoy j

eoy = end of year

eoy n

Figure 6-16: Calculation of outstanding amount

If we continue until the last period, the balance at the end of year n is:

1 2 ...n n nnB P q a q a q a q a− −= ⋅ − ⋅ − ⋅ − − ⋅ − =

{ }2 11 ...n nnB P q a q q q−= ⋅ − ⋅ + + + +

We again see the geometric series in the brackets (see: Section 6.2.1; Page 89), and can therefore simplify the formula to:

On the other hand, the balance of the outstanding amount must be zero at maturity ( ):

We can simplify this term by replacing and, instead of

multiplying the numerator, can divide the denominator by :

{ }2 1 11 ...

1

nn n n

nq

B P q a q q q P q aq

− −= ⋅ − ⋅ + + + + = ⋅ − ⋅−

0nB =

1 10 .

1 1

nn n

n n

q qB P q a a P q

q q

− −= = ⋅ − ⋅ → = ⋅− −

1q i− =nq

1.

1 1 1n

n n n

n

q i P ia P q P

q q q

q

−− ⋅= ⋅ = ⋅ =− − −

6 . 3 L o a n s P a g e | 109

INSTALMENTS

The principal P will be amortised in instalments

payable for n periods if the interest rate is i (compounded periodically).

The periods might be years, months, or quarters; the interest rate i (interest factor ) should be adjusted to the period.

EXAMPLES

1. The loan EUR shall be amortised within 9 periods at an interest rate of 2%. There are no fees. What will the instalments be?

EUR

Cross check

The sum of all payments are 9 times the instalment = 5,513.19 EUR. The principal plus 10% interest is about this amount, which makes sense.

2. A loan EUR should be repaid within 6 months. The effective interest rate is . There are no fees. What will

the instalments be?

The solution to this problem has two parts:

a) We have to calculate the monthly compound interest rate

resulting from the given effective interest rate (see: Section 6.1.3; Page 77):

1 n

P ia

q−⋅=

−

1q i= +

5000P =

5000; 0.02; 1.02; 9 : ?P i q n a= = = = =

9

5000 0.02612.58

1 1.02a −

⋅= =−

4000P =15%effi =

mi


b) Keeping this monthly interest rate in the calculator memory, we can continue:

Cross check

The sum of all repayments is EUR 4,165.60; EUR 4,000 principal repayment and EUR 165.60 interest for half a year at the effective interest rate of 15% makes sense.

Calculation of the instalment contains again the four well-

known financial entries:

• the future value: instalment a

• the present value: principal P

• the time: number of periods n

• the interest rate: rate i, or compound factor q, respectively

The old rule remains true: Three must be known in order to calculate the fourth. Of the above, we already discussed the first bullet.

If we want to calculate the principal P given the instalment, the time and the interest rate, we have to solve the formula for P:

1212 121 1 1 1.15 1 0.0117m eff effi q i= − = + − = − =

4000; 6; ; 1 : ?m m mP n i q i a= = = + =

6

4000694.27

1m

m

ia

q −⋅= =

−

1 n

P ia

q−⋅=

−

(1 )na qP

i

−⋅ −=

6 . 3 L o a n s P a g e | 111

EXAMPLE

Anna can afford monthly instalments of EUR 300 for 12 months. She has been informed by the bank that the monthly compounding interest rate is 2%. Now she can calculate how much she could borrow:

EUR

Cross check

Anna will pay EUR 3,600 in total. This sum contains the principal and the interest. Therefore we have to subtract roughly 28% interest (effective!) from the average of EUR 1500, i.e. about EUR 420. This matches the result almost perfectly.

In the third bullet we solve the equation for time n, because the other three items are given:

In order to isolate n we have to apply the logarithmic operation to both sides:

ln ln ln( )n na P i aq q n q a a P i

a a P i− − ⋅= ↔ = → ⋅ = − − ⋅

− ⋅

ln ln( )

ln

a a P in

q

− − ⋅=

ATTENTION

The logarithm in the numerator may not be split into the difference of the two arguments!

300; 12; 0.02; 1.02 : ?a n i q P= = = = =12300 (1 1.02 )

3172.600.02

P−⋅ −= =

1 11

n nn

P i P i P i a P ia q q

a a aq− −

−⋅ ⋅ ⋅ − ⋅= → − = → = − =

−

ln( )a P i− ⋅


EXAMPLE

How long does Anna have to repay the principal of EUR 4000 if she can afford instalments of EUR 250 every month at n compound interest rate of 1.5% monthly?

months

Cross check

The sum of all payments is 18 times the instalment = 4,600; the principal plus EUR 600 interest for the average outstanding amount of EUR 2000 for 1.5 years at an effective rate of 20% is sufficient.

The fourth bullet would involve calculating the interest rate from the other three financial items. Again this is unsolvable for us, because it requires solving a polynomial equation of degree n, which is beyond the scope of this course.

OUTSTANDING BALANCE

The discussion of instalments and their calculation provided a convenient side effect. Reviewing the process of deriving instalments, we calculated the outstanding amounts for successive periods until we finally reached maturity with the outstanding balance of 0 (see: Figure 6-16; Page 108). In period j the outstanding amount was calculated as:

4000; 250; 0.015; 1.015; ?P a i q n= = = = =

ln 250 ln(250 4000 0.015)18.43

ln1.015n

− − ⋅= =

2 1 1 1(1 ... )

1

j jj j j j

jq q

B P q a q q q P q a P q aq i

− − −= ⋅ − ⋅ + + + + = ⋅ − ⋅ = ⋅ − ⋅−

6 . 3 L o a n s P a g e | 113

OUTSTANDING AMOUNT

The outstanding amount of the principal P at the end of period j is:

with the compounding interest rate i

We should realise that we first have to calculate the instalment a before we can calculate the outstanding balance of the loan in question.

EXAMPLE

A loan of EUR 15,000 will be repaid within 18 months at an interest rate of 1.5% compounded monthly. What is the outstanding amount after 4 (8, 12) months?

EUR

EUR

EUR

Cross check

The total payments within 4 months are four times the instalment, i.e. EUR 3,840. This sum contains effective interest of 20%. The average deposit for the first 4 months is EUR 13,500, which includes interest of roughly 7% for 4 months: EUR 940. Therefore the principal repayment is about 3,840 – 940 = 2900 EUR. That is a relatively precise estimate of the result.

Afterwards, the principal repayments increases because the outstanding balance decreases:

1jj

jq

B P q ai

−= ⋅ − ⋅

15000; 18; 0.015; 1.015: 957.09€; ?1

jn

P iP n i q a B

q−⋅= = = = = = =

−

44

41.015 1

15000 1.015 12005.100.015

B a−= ⋅ − ⋅ =

88

81.015 1

15000 1.015 8826.430.015

B a−= ⋅ − ⋅ =

1212

121.015 1

15000 1.015 5452.700.015

B a−= ⋅ − ⋅ =


The principal repayment within the first 4 months:

15,000 – 12,005.10 = 2,994.90 EUR

The principal repayment during the next 4 months:

12,005.10 – 8,826.43 = 3,178.67 EUR

The principal repayment during the next 4 months:

8,826.43 – 5,452.70 = 3,373.73 EUR

The results are comprehensible.

6.3.3 Effective Interest Rate of Loans

A consumer who wants to take out a loan may be faced with several offers from competing financial institutions. It is therefore of considerable importance to compare the various offers, especially if they are based on different assumptions. The only way to make such comparisons is to use the effective interest rate.

If we want to compare investments that have different compounding periods, we need to state their interest using some common or standardised form. This leads us to distinguish between the nominal interest rate and the effective interest rate effi . This was discussed in

Section 6.1.3 (see: Page 72).

Basically, there are three factors that can create a difference between the effective rate and the nominal rate:

1. The compounding period may be less than or greater than a year. Since the effective interest rate is always an annual rate, it generally differs from the nominal rate.

2. The nominal interest rate may change during the contract time. In this case, the effective interest rate is the average interest rate for the entire contract time when compounding each year. “Average” does not mean that the average is calculated as the mean of the interest rates!

3. Bank charges for services are not covered by the interest rate. These fees are usually collected once, generally at the beginning

6 . 3 L o a n s P a g e | 115

of the contract, sometimes as a fixed cost, sometimes depending on the amount (of a loan, for example).

Often we will see a combination of all three factors.

The effective interest rate is always calculated from the client’s point of view. In Germany, it is set in the law that the effective interest rate must be explicitly stated on every contract involving interest rates. As can be seen, calculating the effective interest rate is a very important topic in banking and finance.

EFFECTIVE INTEREST RATE

The effective interest rate effi is always the annual interest rate taking

into account costs.

We will divide the discussion by looking separately at each of the three aforementioned factors affecting the interest rate.

PERIODIC INTEREST RATE

Compounding in periods of less than a year means that interest is paid out within the year and additional interest is then earned on previously earned interest. We have already discussed this in Section 6.1.3 (see: Page 72). The following paragraph is just a reminder.

If the interest on deposit D is compounded periodically with the nominal interest rate i, the balance of the account after p periods (= one year) will be:

1 (1 )pip

B D= ⋅ +

Since the effective interest rate is always the annual interest (paid or earned), we have to determine how much interest must be paid yearly in order to reach the same balance:

1(1 ) (1 ) (1 ) 1p pi ieff effp p

D i B D i⋅ + = = ⋅ + → = + −

This formula remains true for deposits and loans.


EXAMPLE

For monthly instalments, p = 12, and a nominal interest rate of i = 24%, or a monthly interest rate of 2%i

m pi = = , the effective interest

rate will be:

12 12(1 0.02) 1 1.02 1 0.2682effi = + − = − = = 26.82%

Cross check

The effective interest rate must be more than the nominal rate, which is the case.

REPAYMENT PLAN AND EFFECTIVE INTEREST RATE

A general method of calculating the effective interest rate in the case of a loan may be applied with reference to the repayment plan. The periodically paid interest rate pi

can always be calculated from the

repayment plan as follows :

all interest paid

all outstanding amountspi = ∑∑

This relation is valid for any period: With p being some portion of a year (p = 12 for monthly instalments, p = 4 for quarterly instalments, etc.), the periodic interest rate can be calculated using the above relationship,

and the effective interest rate will be: (1 ) 1peff pi i= + − .

We will illustrate this procedural method by means of three Examples.

6 . 3 L o a n s P a g e | 117

EXAMPLES

1. A typical repayment plan for a loan of EUR 1000, with six monthly instalments of EUR 184.60 and an interest rate of 3%pi = , is:

p OA(p) Int(p) PR(p) Inst(p) Months EUR EUR EUR EUR

1 1000.00 30.00 154.60 184.60

2 686.17 25.36 159.24 184.60

3 522.15 20.59 164.01 184.60

4 353.22 15.66 168.93 184.60

5 179.22 10.60 174.00 184.60

6 0.00 5.38 179.22 184.60

Sum 3586.17 107.59 1000.00 1107.59

Table 6.5: Repayment plan with monthly instalments2

The interest paid per period (= month) is: 107.59

0.033586.17pi = =

This is of course not surprising, because the instalment was calculated with a monthly interest rate of 3%! However, the calculation shows that the approach is correct in the above case.

In order to calculate the effective interest rate, which is always the annual effective rate, we would have to apply the above formula:

12(1 ) 1 1.03 1 42.58%peff pi i= + − = − =

2. A loan of EUR 10000 has a maturity of 7 quarters and a quarterly interest rate of 8%pi = . The quarterly instalments will be with

P=10000, i=0.08 → q=1.08, and n=7:

7

10000 0.081920.72

1 1 1.08n

P ia

q− −⋅ ⋅= = =

− − EUR

The repayment plan for 7 quarters is:

2 Headlines: see page 103


p OA(p) Int(p) PR(p) Inst(p) Quarters EUR EUR EUR EUR

1 10000 800.00 1120.72 1920.72

2 8879.28 710.34 1210.38 1920.72

3 7668.90 613.51 1307.21 1920.72

4 6361.69 508.94 1411.78 1920.72

5 4949.91 395.99 1524.73 1920.72

6 3425.18 274.01 1646.71 1920.72

7 1778.48 142.28 1778.48 1920.76

Sum 43063.45 3445.08 10000.00 13445.08

Table 6.6: Repayment plan with quarterly instalments3

The interest per quarter is: 3445.08

8%43063.45pi = = , i.e. again exactly the

rate we started with, which didn’t change.

The effective interest rate in this case is:

( )4 41 0.08 1 1.08 1 36.05%effi = + − = − =

In the third example we drop the assumption that the interest rate is constant during the contract.

3. We assume that the loan from EUR 1000 with a six-month maturity, the instalments of monthly EUR 185 include the following floating interest payments:


6 . 3 L o a n s P a g e | 119


1 1000.00 35.00 155.00 190.00

2 845.00 30.00 160.00 190.00

3 685.00 25.00 165.00 190.00

4 520.00 20.00 170.00 190.00

5 350.00 17.00 173.00 190.00

6 177.00 13.00 177.00 190.00

Sum 3577.00 140.00 1000.00 1140.00

Table 6.7: Repayment plan with floating interest rate4

The interest rate per month is: 140

0.039143577pi = =

In order to get the effective interest rate, we have to calculate the yearly compounded rate:

( ) 121 1 1.03914 1 58.52%p

eff pi i= + − = − =

A relatively small and seemingly harmless looking change of the interest generated a substantially different effective interest rate.

Cross check

The interest rate for the first month is 3.5%; however, for the last month the interest rate raised to 7.3%. An average rate of about 4% seems plausible.



FLOATING INTEREST RATES

The interest rate may change during the contract time, as in the above repayment schedule. In another example, it is possible to imagine an investment with increasing interest rates. In Germany, there are treasury bonds which are sold exactly under these conditions.

On the credit side, we can also imagine that there could be floating – generally increasing – interest rates, especially if the maturity of a loan is long-term. We start by discussing the less complex problem of repaying a loan in one single payment. In Fig. 6-17, interest ji will be paid in year j

for the principal P, and the principal and all interest, together R, will be repaid at the end of the agreement.

P

eoy 1t

today

R

1q 2q 3q 4q

eoy 2 eoy 3 eoy 4

eoy = end of the year Figure 6-17: Floating yearly interest rates

The corresponding interest factor is 1j jq i= + , and the repayment R −

from the bank's point of view − will be:

1 2 3 4R P q q q q= ⋅ ⋅ ⋅ ⋅ after four years

The effective interest rate is the true rate for the customer. From his (or her) point of view, the repayment will be the investment compounded effectively four times:

4 4(1 )eff effR P q P i= ⋅ = ⋅ +

The repayment is the same for both sides, thus:

41 2 3 4 (1 )effR P q q q q P i= ⋅ ⋅ ⋅ ⋅ = ⋅ +

6 . 3 L o a n s P a g e | 121

The effective interest rate will be independent of the principal amount.

EFFECTIVE INTEREST RATE WITH FLOATING INTEREST

The effective interest rate for floating nominal interest rates is:

41 2 3 4 1effi q q q q= ⋅ ⋅ ⋅ −

EXAMPLE

Let the interest rates be: 1 3%i = for the first year, and 2 5%i = ,

3 6%i = , 4 7%i = , and 5 10%i = for the following years. The effective

interest rate for a loan over five years is:

51.03 1.05 1.06 1.07 1.1 1 0.0617 6.17%effi = ⋅ ⋅ ⋅ ⋅ − = =

Cross check

The arithmetic mean of the five interest rates would be 6.2%. The result is convincing.

In the case of floating interest rates for periods shorter than a year, we have to combine the latter procedure with the first. The year will be replaced by the more general term “period”:

P

eop 1

t

today

R

1q

eop j eop n-1

jq nq......

...

eop = end of period

Figure 6-18: Floating periodical interest rates


Using the same argumentation as before we will obtain a single periodic interest factor:

1 2 ...np nq q q q= ⋅ ⋅ ⋅

Since the effective interest rate is always the true yearly rate, and assuming that the year is divided into p periods, we have to compound the periodic rate p times:

(1 ) 1peff pi i= + −

EXAMPLE

The monthly interest rates for a loan with 6 months maturity are assumed to be: 2%, 2.5%, 3%, 3.3%, 3.8%, and 5%.

What is the effective interest rate?

The average monthly factor will be:

6 1.02 1.025 1.03 1.033 1.038 1.05 1.0326mq = ⋅ ⋅ ⋅ ⋅ ⋅ =

The effective interest rate is:

12 1 0.4699 46.99%eff mi q= − = =

Cross check

Again, we calculate the arithmetic mean of the interest rates, obtaining 3.27% per month. The exact result is very close.

FEES

Almost all banks charge fees when disbursing a loan, some even more than once. Usually it is either a fixed percentage of the loan amount or a fixed amount, paid once at the beginning of the contract. The fee f is basically an in-payment by the customer (see: Fig. 6-19); it is deducted from the principal amount P to be disbursed, so that the customer gets only P− f.

6 . 3 L o a n s P a g e | 123

We start by assuming that the principal plus interest will be repaid after one year in one single payment R.

interest factor per year qnom=1+i

R

ttodayP-f

f

1 year

Figure 6-19: Fees and one repayment

The bank demands nominal interest for the full loan:

(1 )nomR P q P i= ⋅ = ⋅ +

However, the customer gets only P−f, yet he repays R; the true interest must therefore be calculated on the basis of this reduced amount:

( ) ( ) (1 )eff effR P f q P f i= − ⋅ = − ⋅ +

Due to the equality of the repayments we obtain:

(1 ) 1effP

i iP f

= + −−

EXAMPLE

Let the loan be EUR 1000 and the fee 2% with a nominal interest rate of 24%. The effective interest rate is:

10001.24 1 0.2653 26.53%

980effi = ⋅ − = =


Cross check

The effective interest rate must be slightly higher than the sum of the nominal rate and fee because the fee is paid one year before.

If the time between disbursement and repayment is less than a year – for instance 4 months – the result of a similar argumentation will be an average interest rate for the corresponding time interval.

The year may be divided into p periods. We obtain an average interest factor per period:

(1 ) (1 )ip p p

Pq i

P f= + = ⋅ +

−

The effective interest rate will be yearly compounded interest rate:

1 (1 ) 1p

p ieff p p

Pi q

P f

= − = ⋅ + − −

EXAMPLE

Let the loan be EUR 1000 and the fee 2% with a nominal interest rate of 24%. Repayment after four months means p=3:

0.243

1000(1 ) 1.08 1.1020

980pq = + = ⋅ =

pi is the true interest rate for four months, effi the effective rate

(annually!):

3 1 0.3384 33.84%eff pi q= − = =

Cross check

Since a 2% fee is paid for considerably less time and the periodical rate has to be compounded, the effective interest rate must be significantly larger than the nominal interest rate.

6 . 3 L o a n s P a g e | 125

Repaying the principal at once is rather rare. Generally, loans will be repaid in the form of instalments. Therefore, we will now discuss the most probable situation: A customer has to pay a disbursement fee and repays the principal in monthly instalments.

For a loan P, with a nominal monthly interest rate mi (interest factor

1m mq i= + ) and maturity after 12 periods, the amount of instalment a

will be (see: Section 6.3.2; Page 108):

121

m

m

P ia

q −⋅=

−

Now we assume that there is a disbursement fee f (see: Fig. 6-20). Instead of in-paying the fee it is usually subtracted from the principal amount which is indicated in the figure by means of the reduced length of the downward arrow.

ttoday

a a a a12 times

1m mq i= +

...

P-f

f

interest factor per month

Figure 6-20: Fees and instalments

Besides the fee, after 12 months the bank receives the principal amount plus interest, which is 12 times the compounded monthly interest.

Bank receives: 12mP q⋅

The customer received in fact only (P−f) for which he paid the effective interest.


Customer pays: ( ) (1 )effP f i− ⋅ + for the year

Balancing both sides leads to:

12 1eff mP

i qP f

= − −

EFFECTIVE INTEREST RATE FEE AND INSTALMENTS

The effective interest rate for the principal amount P with a monthly interest factor mq , a fee f , and 12 months maturity is:

12 1eff mP

i qP f

= − −

EXAMPLE

P = 1000 EUR; fee = 2%; monthly interest 2%pi = ; n = 12 months

maturity.

The instalments are: a = 94.56 EUR

The effective interest rate is: 1210001.02 1 29.41%

980effi = ⋅ − =

Cross check

Without a fee, we remember that the effective interest rate in the case of monthly compounded interest of 2% was about 26%. Due to the fee, the effective interest rate must be significantly higher. Almost 30% sounds reasonable.

If maturity is less than a year, which is often the case for micro-loans, the effect of a fee will be correspondingly higher.

We now assume that the maturity of the loan is t months. Hence,

6 . 3 L o a n s P a g e | 127

• t < 12 → maturity is less than one year,

• t = 12 → maturity is exactly one year, and

• t > 12 → maturity is more than one year.

After t months the bank receives: tmP q⋅

The customer received only (P−f) and pays for that amount ( ) tP f i− ⋅ .

Hence the customer pays: ( ) (1 )tP f i− ⋅ +

Balancing both amounts leads to:

1 tt t m

Pq i q

P f= + = ⋅

−

This is the interest rate which the customer truly has to pay for the t months. The equivalent interest factor per month is therefore (see: Section 6.1.3; Page 77):

( ) ( ) ( )1 11

1 .t tt tP P Pt tm t t m m mP f P f P fq q q q q q− − −= = = ⋅ = ⋅ =

This is the true monthly compounding factor. In order to obtain the effective interest rate, we have to compound this factor 12 times:

( )12

12 1tPeff mP fi q−= ⋅ −

The effective interest rate for a loan P with a fee f, t months maturity, and monthly compounding interest rate mi is:

( )12

12 1tPeff mP fi q−= ⋅ − , where 1m mq i= +

EXAMPLE

Let us now keep all the conditions the same, except that the loan has to be repaid in t = 6 instalments, i.e.:

P = EUR 1000; f = 2% = 20; 1.02mq = ; maturity t = 6 months


Although we don’t need it for calculating the effective interest rate, bear in mind that repayment in less time certainly will imply a larger instalment: a = 178.53 EUR


( )126 121000

980 1.02 1 0.3205 32.05%effi = ⋅ − = =

Cross check

Compared to the last example, the only change is the shorter maturity. The same fee is therefore charged for less time and the effective interest raises to 32% compared to 29.4%.

The same formula holds for t > 12, i.e. for maturity of more than one year.

EXAMPLE

All data remain unchanged, except that the loan has to be repaid in t = 18 instalments, i.e.:

P = EUR 1000; f = 2% = 20; 1.02mq = ; maturity t = 18 months

Instalment: a = EUR 66.70


( )1218 121000

980 1.02 1 0.2854 28.54%effi = ⋅ − = =

Cross check

Compared to the two previous examples, maturity is now more than one year. The same fee is therefore charged for more time and the effective interest decreases, which was expected.

6 . 3 L o a n s P a g e | 129

EXERCISE 6.3: LOANS

1. Anna receives a loan of EUR 500. She pays it back together with the interest after 6 months. The interest rate is i = 1% compounded monthly. How much must she pay after 6 months?

2. Anna can pay EUR 200 after 2 months (without any fee) and the remaining amount together with the interest at the end of the sixth months. How much does she have to pay?

3. For a loan of EUR 1800 Anna agrees on the following repayment plan: payment of EUR 400 after 3 months, EUR 300 after 6 months, EUR 500 after 8 months, and the remaining amount together with the interest after 12 months. The interest rate is 1% compound monthly. How much is the last payment?

4. Anna receives a loan of EUR 5,000.

a) Calculate the instalment at 1.5% monthly compounded interest and a 12 month maturity.

b) Set up the repayment plan for Anna.

5. Set up the repayment plan for a loan of EUR 2,000 and an effective interest rate 16%, monthly instalments, and a 9 month maturity.

6. Anna receives a loan of EUR 3,000. The monthly interest rate is 1.2%. She wants to repay the loan in 8 equal instalments. Calculate the instalments and the effective interest rate.

7. How long must Anna repay the loan from Problem 6, if she can afford only EUR 250 per month?

8. Anna's financial situation allows her to repay EUR 350 for a loan. Procredit offers her 1.2% monthly interest, if she repays the loan in 12 instalments. What size loan can she get?


9. What is the outstanding balance of Anna's loan from Problem 4 after 6 months?

10. Calculate Anna's outstanding balance from Problem 8 after 4 months.

11. What is the effective interest rate Anna has to pay for the loan in problem 1?

12. Anna asks for a loan of EUR 1400 at a competitor of ProCredit and was offered the follwing conditions:

• Bi-monthly instalments, i.e. every two months.

• 3% interest per period (i.e. for two months) on the principal amount

• 7 periods maturity (= 14 months)

p5 OA(p) Int(p) PR(p) Inst(p) Months EUR EUR EUR EUR

1 1400.00 42.00 200.00 242.00

2 1200.00 42.00 200.00 242.00

3 1000.00 42.00 200.00 242.00

4 800.00 42.00 200.00 242.00

5 600.00 42.00 200.00 242.00

6 400.00 42.00 200.00 242.00

7 200.00 42.00 200.00 242.00

Sum 5600.00 294.00 1400.00 1694.00

5 p=periods; OA(p)=outstanding amount in period p; Int(p)=interest to be paid in period p; PR(p) = principal repayment to be paid in period p; Inst(p) = instalment to be paid in period p

6 . 3 L o a n s P a g e | 131

Interest of 3% for two months signals a nominal rate of 18% or an effective rate of 19.4%. But, is this true?

What will be the effective interest rate of this agreement, really?

13. On 1. January 2009 Anna got a loan of 2000 EUR under the following conditions:

• ProCredit has agreed that she will repay the principal and all interest at once at the end of the fourth year.

• However, Anna has to pay a service fee of 100 EUR when the loan is disbursed.

• For the first year Anna has to pay 8% interest, the next year 9%, the third year 11%, and the last year 15%.

a) How much has the Anna to pay at the end of year 2012?

b) What is the effective interest rate for Anna?

14. Anna receiced a loan of 2000 EUR . She has to repay the principal in 15 months with a monthly compounding interest rate of 1.5%. At the time of disbursement she has to pay a fee which is 3% of the principal.

a) Calculate the instalments.

b) Prepare the repayment plan for Anna.

c) What is Anna’s effective interest rate?


ANSWERS 6.3: LOANS

1. Repayment is: R = 530.76 EUR

2. Repayment is: R = 322.64 EUR

3. Last Payment: R = 752.05 EUR

4. a) Instalment: a = 458.40 EUR

b) Repayment plan6 :


1 5,000.00 75.00 383.40 458.40

2 4,616.60 69.25 389.15 458.40

3 4,227.45 63.41 394.99 458.40

4 3,832.46 57.49 400.91 458.40

5 3,431.55 51.47 406.93 458.40

6 3,024.62 45.37 413.03 458.40

7 2,611.59 39.17 419.23 458.40

8 2,192.36 32.89 425.51 458.40

9 1,766.85 26.50 431.90 458.40

10 1,334.95 20.02 438.38 458.40

11 896.58 13.45 444.95 458.40

12 451.63 6.77 451.63 458.40

6 Headline: see footnote page 128

6 . 3 L o a n s P a g e | 133

5. Instalment is: a = 236.28 EUR

Repayment plan7:


1 2,000.00 24.89 211.39 236.28

2 1,788.61 22.26 214.02 236.28

3 1,574.59 19.60 216.68 236.28

4 1,357.91 16.90 219.38 236.28

5 1,138.53 14.17 222.11 236.28

6 916.41 11.40 224.88 236.28

7 691.54 8.61 227.67 236.28

8 463.87 5.77 230.51 236.28

9 233.36 2.90 233.38 236.28

6. Instalment a = 395.53 EUR; effective interest

7. Possible maturity: n = 13 months

8. Possible principle is: P = 3,889.95 EUR

9. Outstanding amount is: EUR


0.1539 15.39%effi = =

6 2611.59B =


10. Outstanding amount is: 4 2654.65 EURB =

11. 121.1 1 0.1268 12.68%effi = − = =

12. True bi-monthly interest rate: 2942 5600

0.0525mi = =

Effective interest rate: 61.0525 1 0.3594 35.94%effi = − = =

13. a) 1 2 3 4 3005.39R L q q q q= ⋅ ⋅ ⋅ ⋅ = EUR

b) The effective rate is: 12.15%effi =

14. a) 15

2000 0.015149.89

1 1 1.015n

P ia

q− −⋅ ⋅= = =

− −EUR

6 . 3 L o a n s P a g e | 135

b) Repayment Plan8


1 2000.00 30.00 119.89 149.89

2 1880.11 28.20 121.69 149.89

3 1758.42 26.38 123.51 149.89

4 1634.91 24.52 125.37 149.89

5 1509.54 22.64 127.25 149.89

6 1382.29 20.73 129.16 149.89

7 1253.14 18.80 131.09 149.89

8 1122.05 16.83 133.06 149.89

9 988.99 14.83 135.06 149.89

10 853.93 12.81 137.08 149.89

11 716.85 10.75 139.14 149.89

12 577.71 8.67 141.22 149.89

13 436.49 6.55 143.34 149.89

14 293.15 4.40 145.49 149.89

15 147.65 2.21 147.66 149.87

Sum 16555.23 248.33 2000.00 2248.32

c) ( )1215 122000

1940 1.015 1 1.2251 22.51%effi = ⋅ − = =



6.3.4 Progress Test “Loans”

You should allot yourself some time for concentrated work on this test. Try to solve as many problems as you can. Do not use the reader to look for the solution. The aim of the test is to get feedback on how much you know or have learned up to now.

At the end of the chapter, you will find the solutions to the problems. Each solution is reduced to the final answer; it may even be only a number, a symbol, a table or a graph. You should check your answers. If they are correct, you may continue and start the next chapter. In any other case (you got the wrong answer or no answer at all) you will find some advice on what to do. In most cases, you will be directed to the section in the reader which you should repeat in order to close the gap.

1. A loan of EUR 800 must be repaid in a single payment after 6 months. The monthly compounded interest rate is 1%. What is the repayment amount?

2. How much would the monthly instalments be for Problem 1?

3. A client wants to compare the offer of a competitor with the conditions at ProCredit for a loan of EUR 2,000 with a 9 month maturity.

The competitor offers 15% nominal interest plus a service charge and calculated monthly instalments of EUR 244.34.

ProCredit asks for 16 % effective interest. What is the instalment amount for the ProCredit loan?

4. Which loan amount can you offer as a loan officer, if the client's monthly instalment should not exceed EUR 140 with an effective interest rate of 15% and a 9 month maturity?

6 . 3 L o a n s P a g e | 137

5. Anna asks for a loan of EUR 1,000 because she wants to buy a notebook immediately. ProCredit offers an interest rate of 1.5%, compounded monthly with a 12 month maturity.


b) Set up the repayment plan.

6. The client from Problem 4 asks for a loan of EUR 2,000. He agrees to monthly instalments of EUR 150. Calculate the maturity of the agreement.

7. A mortgage loan of EUR 200,000 should be repaid within 20 years at an effective interest rate of 8%.

a) Calculate the annuity.

b) What is the outstanding amount after 10 years?

8. Anna and ProCredit signed the following savings plan:

• Anna deposits EUR 50 every month.

• The monthly compound interest rate rises from 1.5% in the first and second year, to 1.8% during the third and fourth year, and to 2.5% during the fifth year.

a) What will be the balance at the end of fifth year?

b) What is the effective interest?

9. Anna receives a loan of EUR 4000 with a nominal interest rate of 18%. Maturity is 18 months, and disbursement fee is 2.5%.


b) Calculate the effective interest rate.


6.4 Answers to Progress Tests

6.4.1 Answers to Progress Test “Interest”

You should check your answers. If they are correct, you may continue and start the next chapter. In any other case (you got the wrong answer or no answer at all), repeat the topics in the corresponding sections in order to close the gap.

1. 141 interest days → 309.40 EUR

2. a) EUR 885.78 b) EUR 904.36 c) EUR 908.80

3. a) 10% b) 10.38% c) 10.47%

4. EUR 651.97

5. 65.92 months

6. 1.046 % monthly

7. a) 3.443% b) 2.282% c) 1.135%

7 . 0 S u m S y m b o l P a g e | 139

6.4.2 Answers to Progress Test “Deposits”

You should check your answers. If they are correct, you may continue and start the next chapter. In any other case (you got the wrong answer or no answer at all), repeat the topics in the recommended sections in order to close the gap.

1. EUR 1,180.70

2. EUR 566.26

3. EUR 1,008.11

4. EUR 47.20

5. EUR 77.92

6. 18.28 months


6.4.3 Answers to Progress Test “Loans”

You should check your answers. If they are correct, you may continue and start the next chapter. In any other case (you got the wrong answer or no answer at all), repeat the topics in the corresponding sections in order to close the gap.

1. EUR 849.22 2. EUR 138.04 3. EUR 236.28

4. EUR 1,185.04 5. a) EUR 91.68

b) Repayment Plan9


1 1,000.00 15.00 76.68 91.68

2 923.32 13.85 77.83 91.68

3 845.49 12.68 79.00 91.68

4 766.49 11.50 80.18 91.68

5 686.31 10.29 81.39 91.68

6 604.92 9.07 82.61 91.68

7 522.32 7.83 83.85 91.68

8 438.47 6.58 85.10 91.68

9 353.37 5.30 86.38 91.68

10 266.99 4.00 87.68 91.68

11 179.32 2.69 88.99 91.68

12 90.33 1.35 90.33 91.68


7 . 0 S u m S y m b o l P a g e | 141

6. 67 months

7. a) EUR 2,0370.44 b) EUR 136,687.32

8. a) EUR 5738.25 b) 24.15%effi =

9. a) EUR 255.22 b) 21.6%effi =

7 . B a s i c s o f S t a t i s t i c s P a g e | 143

7. Basics of Statistics


Statistics




4. Basic Functions


1. Introduction

Par

t 2: A

dvan

ced

Prerequisites: All calculations in statistics depend on the correct application of algebra. Therefore it is necessary to be able to deal confidently with the transformation and evaluation of expressions and equations.

144 | P a g e 7 . B a s i c s o f S t a t i s t i c s

Learning Targets: Statistics is the art of making numerical conjectures about puzzling questions:

• What are the effects of increasing the interest rate?

• How can a class’s exam performance be measured?

• How much profit does a casino make on its roulette tables?

• How many job applicants are actually hired?

• What is the inflation rate in a given country?

These are complex issues, and statistical methods can be very useful for analysing them. Statistical methods have been developed over hundreds of years, and weighty volumes have been written about them.

Here we will study just some of the basic statistical methods of descriptive statistics.

WHAT IS STATISTICS ?

The word is Latin in origin (status → state, condition, situation, position, shape): It incorporates all methods of analysing large sets of data by summarising them in the form of concentrated information. Instead of analysing all data – which might be very time consuming – we summarise the data by classifying them in tables, representing them in graphs, and creating key figures which reveal the basic properties of the set of data. Of course, in each of these steps we lose some of the details contained in the complete set of information, yet at the same time, by summarising the data, we gain a better overview of the situation as a whole. Key figures provide a much better insight into the conditions reflected in the data. The shape of a column diagram or a curve can be recognised at a glance. Instead of looking at the details we are able to see the “big picture”. This is precisely what the concentration of information is about: it is the intended purpose of statistics.

7 . B a s i c s o f S t a t i s t i c s P a g e | 145

Statistics can be subdivided into descriptive and conclusive statistics. The former is the art of summarising and concentrating data. The latter tries to deduce properties from a sample of the complete set of data and draw conclusions regarding the whole using probability theory. In this course we focus on the fundamentals of descriptive statistics. We will discuss the collection, tabulation and systematic classification of quantitative data with the aim of gaining an overview of the set of data in order to draw conclusions about future trends. We will discuss the concentrated representation of large sets of data in tables and in diagrams. Graphical representations are ideal tools for providing a general picture of the situation as a whole, which can then be easily visualised and analysed at a glance. That is why they are frequently used in presentations, lectures, speeches or academic papers. Summarising data by calculating key figures, such as the average, minimum, maximum and median of the list of observed numbers, is probably the most prominent part of descriptive statistics. These numbers provide us with information about the concentration, i.e. the focus or the centre of gravity, of the data. The final topic under discussion here will be the spread of the data. While the average value identifies the centre point of the data, the variance provides information on how widely or narrowly the data are spread or distributed around the centre. Only if we have both key figures, the centre and the variance, will we have a complete picture of the whole set of data.

Statisticians usually have to handle huge amounts of data. If one were to do everything manually – the calculations, the concentrations of data, and the representations in suitable forms – the work would become tedious, monotonous and tiresome. This is especially true of all banking operations which are naturally based on large sets of data (account, balances, loans, deposits, etc.). The use of suitable software is indispensable in these areas. We will use Microsoft Excel, which offers a broad variety of useful statistical functions and can generate nice-looking diagrams. The screenshots in this chapter were made with the version of Excel that forms part of MS Office 2007.


ESTIMATION

Modern statistics distinguishes between the population and a sample. A sample may be used to gain certain information about the – in its totality – unknown population from which it was drawn. For the most part, the data which mathematical statistics deals with consist of random samples constructed in such a way that every item in the population has an equal chance of being chosen in the sample.

In this course, in line with the needs of the ProCredit group, we will mainly focus on descriptive statistics, i.e. the branch of statistics devoted to summarising the large sets of data that make up the population. This means that we will not discuss any probability-based statistical problems, nor will we argue with samples. However, in some very rare cases – when it comes to discussing variance, for example – we will have to differentiate between population and sample. We will return to the distinctions that need to be made in these cases when discussing the topics that require us to be aware of them.

In economic statistics we generally do not know everything that we would like to know about the population from which the data are a sample. Therefore, we have to consider the data as a random sample of the hypothetically infinite population. For example, if a set of prices for some banking service is recorded it must be considered as a random sample of all (unrecorded) service prices charged by all the other banks in the market.

Statistical estimation is the method by which we extract information about the population from the sample. This is the only object of sampling. The estimates derived from the sample are functions of various observations made on the sample. Since each observation is subject to a certain probability, they are random variables. Consequently, the estimates themselves are random variables as well.

One important property of estimation is that it is unbiased. An unbiased estimate is defined as an estimate whose mean value is equal to the corresponding value of the population. Since we usually use the population for our statistical analysis, the distinctions between biased and unbiased data are not really important for us.

7 . 0 S u m S y m b o l

7.0 Sum Symbol

In statistics we often have to add quite regularly have to describe these sums of entries. In order to avoid writing the complete sum, mathematicians use a short notation for sumsThe so-called sum symbol is simply the Greek capital sigma:it is used quite extensively in this chapter it is defined and explained in this section.

In order to define how many terms should be addedsymbol is usually accompanied by a count index (often one of the letters i, j, k) with its starting and ending value. The starting value is written under the sigma, while the ending value is given on top:


7.0 Sum Symbol

7.1 Presentation Techniques

Tables

Diagrams

Sum Curves

7.2 Key Figures:

P a g e | 147

In statistics we often have to add up large quantities of numbers, and we quite regularly have to describe these sums of entries. In order to avoid

mathematicians use a short notation for sums. is simply the Greek capital sigma:Σ . Because

it is used quite extensively in this chapter it is defined and explained in

In order to define how many terms should be added together, the sum symbol is usually accompanied by a count index (often one of the letters

) with its starting and ending value. The starting value is written under the sigma, while the ending value is given on top:


7.0 Sum Symbol

7.2 Key Figures: Centre

Mean

Median

Mode

7.3 Key Figures: Spread

Variance

Standard Deviation

Normal Curve


The notation 10

1j=∑ indicates that the sum has 10 terms.

The sum symbol is nothing more than an operator telling us that we have to take the sum of a certain number of terms. The term is the operand following the symbol. Generally it contains the count index j; however, the term may also be independent of j (see problem 4 in the examples below).

Very often the term is simply one variable with index j. In this case we get:

1 21

... ...n

j j nj

x x x x x=

= + + + + +∑

However, the term may also be any algebraic term, containing the count index as a parameter and/or index:

30 1 2 3

1 2 30

jj o

j

a x a x a x a x a x=

⋅ = ⋅ + ⋅ + ⋅ + ⋅∑

Thus, we can write a general polynomial expression of degree 3 (see Volume 1, Section 2.3.1) using the sum symbol.

SUM SYMBOL ∑

The ( )n

j k

t j=∑ is a short notation for the sum of terms ( )t j where j has to

be replaced by all consecutive integers starting from j=k (the starting value or lower bound) and ending at j=n (the ending value or upper bound). The starting value k cannot be greater than the ending value: k n≤

7 . 0 S u m S y m b o l P a g e | 149

EXAMPLES

1. The sum of all natural numbers from 1 to 5 can be written using the sum symbol:

5

1

1 2 3 4 5j

j=

+ + + + =∑

2. The sum of the odd numbers from 1 to 21 can be denoted as:

11

1

1 3 5 ... 19 21 (2 1)j

j=

+ + + + + = −∑

Cross check

We find that the first term will be 2 1 1− = , and the last term in the sum is 22 1 21− = . Therefore, the representation with the sum symbol is correct.

3. The geometric series of n terms with quotient q can also be written using the sum symbol (see Section 6.2.1; Page 89):

12 1

0

1 ...n

n j

j

q q q q−

−

=+ + + + =∑

ATTENTION

The sum has n terms if the counter index starts with 1 and ends with n. However, if the counter index begins with j = 0 there is one term more. Therefore, the last sum has n terms even though the counter index ends with n-1.

4. The sum 8

1

1j =∑ might look rather strange at first glance; however,

once you write the complete sum and apply the rules consistently the above expression makes sense:

8

1

1 1 1 1 1 1 1 1 1 8j=

= + + + + + + + =∑


CALCULATIONS USING THE SUM SYMBOL

You have to follow some very strict rules if the sum symbol is to be used in algebraic calculation. Always remember and take care to ensure that

• the sum symbol replaces brackets.

The impact of the sign is to place the entire sum inside a pair of implicit brackets. This is extremely important if multiplicative factors are involved:

( )3

1 2 31

jj

c x c x x x=

⋅ = ⋅ + +∑ or

( )3

1 2 31

jj

c x c x c x c x=

⋅ = ⋅ + ⋅ + ⋅∑

The fact that the sum sign implicitly replaces brackets means that the range of that rule ends with

• the term losing uniqueness when it is composed without brackets.

For instance:

32 2 2 2

1 2 31

jj

x b x x x b=

+ = + + +∑ , but

32 2 2 2 2 2 2

1 2 3 1 2 31

( ) 3jj

x b x b x b x b x x x b=

+ = + + + + + = + + +∑

The brackets in the second expression are necessary if b should be part of the summation.

A similar situation occurs if parts of the sum

• are to be factored out.

If a symbol (number or variable) appears as a factor in every summand it can be factored out. Instead of putting it in front of the brackets (replaced by the sum symbol) it can be taken out of the sum and placed before the sum sign:

7 . 0 S u m S y m b o l P a g e | 151

1 1

n n

j jj j

c x c x= =

⋅ = ⋅∑ ∑

If the term of the summation is a sum itself

• the sum symbol and additive operators can be interchanged.

This means that we can split the term of the sum symbol into individual sums.

2 2 2

1 1 1 1 1 1

( 2) 2 2n n n n n n

j j j j j jj j j j j j

x x x x x x n= = = = = =

+ + = + + = + +∑ ∑ ∑ ∑ ∑ ∑


EXERCISE 7.0: SUM SYMBOL

1. Simplify the expressions using the sum symbol and a count index:

a) 10 20 30 40 50+ + + + =

b) 1 2 10+ + =…

c) 3 6 9 12 15 27 30+ + + + + + + =…

d) 6 7 19 20x x x x+ + + + =…

e) 1 1 1 1 1 1

12 3 4 5 99 100

+ + + + + + + =…

2. Write the sums given in symbolic form in explicit terms:

a) 7

2 13

jj

x −=

=∑ b) 2

22

jj

x=−

=∑

c) 2 15

1

j

j

j x −

=⋅ =∑ d)

3

0

jj

j

a x=

⋅ =∑

e) 3

21

j

j

x

j=

=

∑ f) 6

2

3

( )jk

x x=

− =∑

g) 10

1

2j=

=∑ h) 7

2

11

2j

x=

− =∑

i) 6

0

j

j

q=

=∑

3. Evaluate the sums as much as possible:

a) 5

2

1

( 4)jj

x=

− =∑ b) 5

1

3( 2) ( 2)j jj

x x=

− ⋅ + =∑

7 . 0 S u m S y m b o l P a g e | 153

c) 2 3

1 1

( 4)j jj j

x x= =

− ⋅ = ∑ ∑

4. Factor out as much as possible:

a) 3

2 3

1

2 j j

j

x=

⋅ =∑ b) 3

1

3 jj

x x=

⋅ ⋅ =∑

c) 5

2

3

(2 4 6)k kk

x x=

+ − =∑


ANSWERS 7.0: SUM SYMBOL

1. Simplify the expressions using the sum symbol and a count index:

a) 5

1

10 20 30 40 50 10j

j=

+ + + + =∑

b) 10

1

1 2 ... 10j

j=

+ + + =∑

c) 10

1

3 6 9 ... 27 30 3j

j=

+ + + + + =∑

d) 20

6 7 206

... jj

x x x x=

+ + + = ∑

e) 100

1 1 1 12 3 100

1

1 ...j

j =+ + + + =∑

2. Write the sums given in symbolic form in explicit terms:

a) 7

2 1 5 7 9 133

...jj

x x x x x−=

= + + + +∑

b) 2

2 4 2 0 2 42

jj

x x x x x x− −=−

= + + + +∑

c) 2 1

53 5 7 9

1

2 3 4 5j

j

j x x x x x x−

=⋅ = + ⋅ + ⋅ + ⋅ + ⋅∑

d) 3

2 30 1 2 3

0

jj

j

a x a a x a x a x=

⋅ = + ⋅ + ⋅ + ⋅∑

e)

2 33

21 4 9

j

j

x x xx

j=

= + +

∑

7 . 0 S u m S y m b o l P a g e | 155

f)

62 2 2 2 2

3 4 5 63

( ) ( ) ( ) ( ) ( )kk

x x x x x x x x x x=

− = − + − + − + −∑

g) 10

1

2 2 2 2 ... 2 20j=

= + + + + =∑

h) 7

2

11

2j

x=

− =∑ not defined

i) 6

2 3 4 5 6

0

1j

j

q q q q q q q=

= + + + + + +∑

3. Evaluate the sums as much as possible:

a) 5 5 5

2 2

1 1 1

( 4) 8 80j j jj j j

x x x= = =

− = − +∑ ∑ ∑

b) 5 5

2

1 1

3( 2) ( 2) 3 60j j jj j

x x x= =

− ⋅ + = −∑ ∑

c)

2 22 2

1 1 2 2 1 21 1

( 4) 2 8 8j jj j

x x x x x x x x= =

− ⋅ = + ⋅ + − − ∑ ∑

ATTENTION

This answer would be wrong:

( )2 2 2 2

2

1 1 1 1

( 4) ( 4) 4j j j j j jj j j j

x x x x x x= = = =

− ⋅ ≠ − ⋅ = − ∑ ∑ ∑ ∑ !


4. Factor out as much as possible:

a) 3 3

3 3

1 1

3 3j j

j j

x x= =

= ⋅∑ ∑

b) 3 3

1 1

3 3j jj j

x x x x= =

⋅ = ⋅∑ ∑

c)

5 5 52 2

3 3 3

(2 4 6) 2 4 18k k k kk k k

x x x x= = =

+ − = + −∑ ∑ ∑

7 . 1 P r e s e n t a t i o n T e c h n i q u e s


Prerequisites: There is no special mathematical knowledge required except your readiness to find out about the best way to represent large sets of gain a qualitative overview of the whole.

Excel provides a wide range of tools which help to generate tables and diagrams. Therefore it is a good idea to study the functions” sectionhelpful to know how different diagrams can be produced using


7.0 Sum Symbol


Tables

Diagrams

Sum Curves

7.2 Key Figures:

7 . 1 P r e s e n t a t i o n T e c h n i q u e s P a g e | 157

Presentation Techniques

There is no special mathematical knowledge required except your readiness to find out about the best way to represent large sets of data in order to gain a qualitative overview of the whole.

provides a wide range of tools which help to generate tables and diagrams. Therefore it is a good

to study the options available in the “statistical section of the program. It is also very

helpful to know how different diagrams can be produced using Excel or similar tools.


7.0 Sum Symbol


Mean

Median

Mode


Variance

Standard Deviation

Normal Curve


Learning Targets: Descriptive statistics is the art of summarising large sets of data.

In all our institutions we are dealing with thousands of data records, such as disbursed loans, maturities, account balances or average transactions. We are interested in summarising the observed data in such a way that the management, our colleagues and we ourselves are able to grasp and understand the main attributes of the complete list. Tables, diagrams and curves are the main instruments for doing so.

7.1.1 Tables

A list of characteristic values or property values forms a statistical sequence. Very often we get lists with just one entry; in this case the attribute values form a one-dimensional list. The attributes may be:

• Nominal: for instance → Name (John, Anna, Juan,...); Marital status (single, married); Gender (female, male); etc.

• Ordinal (can be ordered by connotation): for instance → grades (A, B, C,...); dress size (S, M, L, XL,...); loan size range (micro-micro, micro, small, medium,…); etc.

• Cardinal (numbers): for instance → temperature (25°C), speed (100 km/h); loan amount (EUR 2500); age (36 years); etc.

In many cases, especially in the banking sector, the attributes are cardinal or ordinal. Therefore we assume throughout this reader that the attribute values can be ordered by connotation or by size, i.e. that they are ordinal or cardinal.

Let us consider an Example the grades attained by 82 students in their final exam in statistics. The list of all students would look like Tab. 7.1, which is not shown in its full length. If we want to know whether or not the class as a whole was successful in the exam, the list with the individual grades does not help us much because it is too detailed. Therefore, we usually start by creating a tally list, which simply means counting the frequencies of the different grades.


No Name Pts Grade No Name Pts Grade

1 Alois 46 D . . . . . . 2 Juan 80 B 78 Willi 113 A 3 David 92 C 79 Jan 78 B 4 Kathrin 108 A 80 Hans 29 F 5 Annette 98 D 81 Rolf 97 A

. . . . . . 82 Pamela 100 A

Table 7.1: List of grades

The result could look like Tab. 7.2, which provides a much better summary of the class’s exam performance. With nine A grades, only 3 Fs (failed), and a relatively large number of Bs and Cs, one could state that the results are encouraging. Thus we can see that the frequency offers a much clearer overall impression than the preceding detailed list.

Grade Frequency Percent Relative Frequency

Sum of Rel. Frequencies

A 9 11.0% 0.110 0.110

B 25 30.5% 0.305 0.415

C 34 41.5% 0.415 0.829

D 11 13.4% 0.134 0.963

F 3 3.7% 0.037 1.000

Sum 82 100.0%

Table 7.2: Frequency table of grades

The frequency table provides a good overview of the exam: We can see at a glance that the overall result is rather good, because good grades are more frequent than poor grades. Therefore, by summarising the long list of individual grades we have achieved a much clearer impression of the quality of the class’s exam performance at the price of losing the details of the individual students’ grades.


For each grade (each line) the percentage is the portion of students with a certain grade relative to all students: “9 out of 82” → 9

8211%=

The relative frequency is basically the same thing expressed as a decimal number: “9 out of 82” → 9

820.110=

For a list of n items (= grades) the frequency of item k is commonly

denoted by kf , and for the relative frequency we use: kk

fh

n=

The percentage is simply the relative frequency multiplied by 100.

An additional source of information is the “sum of relative frequencies” (the last column of Tab. 7.2). This is the sum of all relative frequencies for all items less than or equal to the item in the corresponding line. However, the sum of relative frequencies only makes sense for cardinal or ordinal attribute values. It can be very useful to know the sum of relative frequencies. For instance, if we know that grades A to D represent a “pass”, the sum of the D grades tells us that 96.3% of all students passed with grade D or better. In the following we usually refer to this curve as the sum curve.

EXCEL : In the past, tally lists of frequencies used to be created very laboriously using pencil and paper. With software like Excel it is now much easier to analyse even very long lists using the function COUNTIF. In the above example, COUNTIF counts the number of cells within a range that meet a given condition. The range would be the list of all grades; the condition would be the corresponding grade. The result is the second column of Tab. 7.2.

GROUPED DATA

Suppose you are the statistician of your bank and you are asked to provide an overview of the number of loans in the loan portfolio. Obviously it would make no sense to use one category for every different principal amount, because there would simply be too many items (= different principals) in the list. Therefore, all of the banks in our group classify their loans into different size categories by dividing the attribute


value into intervals from “micro-micro” to “large” and assigning each loan to the appropriate category.

In the following Tab. 7.3 we illustrate10 the classification scheme by numbers. For the purposes of illustration, we have chosen to use only four groups: Micro, Small, Medium and Large. “Left” (“Right”) denotes the left (right) bound of the corresponding group in EUR thousands. Thus, “Micro” ranges from 0 to EUR 4000, while “Medium” stands for loans between EUR 10,000 and EUR 30,000. “Middle” refers to the middle point in the corresponding interval.

The frequency in the category “Micro” means that we identified 377 loan contracts with disbursed amounts smaller than or equal to EUR 4000. The relative frequency shows that these make up 13.4% of all loans, and the sum of relative frequencies tells us that, for instance, 87.4% of all loans are smaller than EUR 30,000.11

If we have to summarise large sets of data with many different attribute values, we have to categorise the list entries into different groups. That means that we must further condense the large volume of information. The grouped frequency tables with grouped data provide a very comprehensive insight into the loan portfolio. However, we gain this at the price of the loss of detailed information about the individual contracts.

No. Class Left Right Middle Freq. Relative Freq.

Sum of Rel. Freq.

1 Micro 0 4 2 377 0.134 0.134

2 Small 4 10 7 1429 0.509 0.643

3 Medium 10 30 20 649 0.231 0.874

4 Large 30 90 60 355 0.126 1.000

Sum 2810 1

Table 7.3: Frequencies of grouped data

10 Neither the numbers nor the scales are realistic. They have been chosen simply in order to create tables and diagrams that clearly illustrate the intended purpose. 11 Again, the figures presented here are not realistic. The magnitudes have been chosen in order to demonstrate the properties of the diagrams in an easily understandable way.


In Tab. 7.3 the number of entries within each group and their relative frequencies are listed. The relative frequency is the number of entries per group divided by the number of entries in the list. The middle point of the group is often used as the representative or mean value.

Let us denote:

k = the count index for the class k

j = the count index for the item j

m = the number of attributes (groups) defining the length of the list

n = the number of items (entries) in the statistical sequence (population)

kf = the frequency of items in group k

kh = the relative frequency of items in group k

FREQUENCY AND RELATIVE FREQUENCY OF GROUPED DATA

Using the above symbols we can conclude:

1

m

kk

n f=

=∑ → The sum of all frequencies is equal to the size of the

population (the length of the list of all items).

kk

fh

n= → The relative frequency is given by the absolute frequency

divided by the size of the population.

1

1m

kk

h=

=∑

EXCEL : Grouped data is a simple but extremely efficient way to aggregate large sets of data and thus concentrate information. However, it is also hard work to go through long lists and count entries of different groups using a tally list. Excel provides a helpful function to count frequencies. In the “Statistical Functions” domain you will


find the function FREQUENCY. However, it is slightly complicated to apply:

• Define the intervals of the groups; they can be equidistant, meaning that the intervals all have the same width. They may also have different widths. The right endpoints are included; for instance, 15 will be in group k = 1 of Tab. 7.5.

• Include the function FREQUENCY in the first position of column “absolute frequency” and define as “Data Array” the list of entries which are to be grouped.

• Define “Bins-Array” by marking the array of Right end points in the class list.

• Pull formula into all class rows and keep them marked.

• Press key F2 and then CTRL+SHIFT+ENTER.

EXAMPLE

In Tab. 7.4 a list of 60 entries is given in the form of a matrix:

25 15 16 30 26 5 20 30 24 37 21 25

25 17 19 25 28 26 52 26 30 26 21 45

49 22 21 43 26 27 19 29 32 27 30 12

66 13 23 29 27 39 29 28 35 27 40 48

57 1 24 28 28 33 27 29 38 26 63 41

Table 7.4: List of data

EXCEL : Assuming that the data array is in the area D6:O10 and the bins-array (the left value of the class range) is in the area S5:S10, the Excel formula in the first line of column “Abs.Freq.” has to be:

=FREQUENCY(D6:O10;S5:S10)


This formula is tracked down the complete table; the column stays marked. Then press F2 and afterwards CTRL+SHIFT+ENTER. The result will be the frequencies

kf in column “Abs.Freq.”. Their sum is n = 60.

Now you can calculate the relative frequencies: kk

fh

n= and the sum of

the relative frequencies "Freq.Sum". The result is shown in Tab. 7.5.

Class Left Right Middle Abs.Freq. Rel.Freq. Freq.Sum

k kl kr km kf kh

1 0 15 7.5 5 0.0833 0.0833

2 15 25 20 16 0.2667 0.3500

3 25 30 27.5 23 0.3833 0.7333

4 30 40 35 7 0.1167 0.8500

5 40 50 45 5 0.0833 0.9333

6 50 70 60 4 0.0667 1.0000

60 1

Table 7.5: Grouped frequencies

Just imagine: In your bank the list of items probably contains several thousand entries. In this case it is very important

1. to summarise the data in suitable frequency tables and

2. to apply Excel in order to compile the list.


7.1.2 Diagrams

Another good way to summarise large sets of data is the graphical representation. Typically one first has to condense the data in frequency tables which then serve as the sources of data for column, line, pie, bar or area diagrams. The purpose of a diagram is to communicate the main features of the information to the addressee. The properties of the complete set of data can be grasped at a single glance.

Let us use as an example the frequency Tab. 7.2 showing the exam results. One way to judge the quality of the students’ exam performance would be to count the excellent and very good grades (A and B) and compare them with the corresponding numbers in other categories (say, D and F). In Fig. 7-1 we have created a column diagram based on the frequency distribution. The heights of the columns communicate a clear picture that the majority of the students achieved good grades because the concentration of the columns towards the good grades (the left side) is obvious.

0

5

10

15

20

25

30

35

40

A B C D F

Results of the Exam

Grades

Num

ber

of S

tude

nts

Figure 7-1: Column diagram of grades

In principle, we can handle tables with the attribute values fragmented in intervals (grouped attribute values) in the same way. The result is shown in Fig. 7-2. Again, we receive a clear picture: Almost 400 loans are in the range 0 – EUR 4,000 (Micro), while about 1,400 loans of between EUR 4,000 and EUR 10,000 (Small) are outstanding. However, Fig. 7-2 differs


from Fig. 7-1 in an important respect. In Fig. 7-1 the columns are assigned to a single attribute only, while in Fig. 7-2 each column represents the whole range of values of the corresponding class.

EXCEL : Excel provides a large set of different diagram templates. When you select Insert Chart you are prompted to choose a template from the list. The chart wizard offers a lot of support for creating the appropriate presentation.

0

2 0 0

4 0 0

6 0 0

8 0 0

1 0 0 0

1 2 0 0

1 4 0 0

1 6 0 0

M ic ro Sm all M e d ium L arge

Loan Portfolio by Class of Loans

Categories

Num

ber

of L

oans

Figure 7-2: Column Diagram of a frequency table

In the above column diagram, each column represents a whole class. Consequently, statisticians often use a slightly modified form of the column diagram.

• They select a representative point for the class, which is in many cases the middle point of the interval.

• They close the gap between the columns, demonstrating that the area of the column represents the frequencies of the group as a whole.

• They divide the height of each column, i.e. the frequency of the class members, by the width of the interval. Thus, the area of the


column, i.e. the height multiplied by the width, becomes equivalent to the portion of the class as a whole.

In Tab. 7.3 we divided the frequencies (column "Freq.") as well as the relative frequencies ("Rel.Freq.") by the width of the corresponding class. In Tab. 7.6 we see the results in columns headed “Freq. Density”, and “Rel.Freq. Densisty”, respectively.

Class L R M Freq. Freq.

Density Rel.

Freq.

Rel. Freq.

Density

Sum of Rel.

Freq.

Micro 0 4 2 377 94.25 0.134 3.35% 0.134

Small 4 10 7 1429 238.17 0.509 5.09% 0.643

Med. 10 30 20 649 32.45 0.231 0.66% 0.874

Large 30 90 60 355 5.92 0.126 0.18% 1.000

Sum 2810 1

Table 7.6: Frequency density of grouped data

Finally, we again produce a column diagram with the relative frequencies (in percentages) as data: Fig. 7-3. This type of diagram is referred to as the density distribution of the relative frequencies of the class attributes. Statisticians call this type of column diagram a histogram. Because the area of each column is exactly equivalent to its portion of the whole, the sum of all areas must be 1.


Density Distribution

Categories

Freq

uenc

y D

ensi

ty

0.0%

1.0%

2.0%

3.0%

4.0%

5.0%

6.0%

Micro Small Me dium Large

Figure 7-3: Column diagram of a grouped frequency table

EXCEL : It is common sense to close the gaps between the different columns in the diagram when representing densities. This is because in the density distribution the entire area under the curve will be normalised to 1. If the gaps are closed, all the columns are connected.

If you want to close the gaps, just click on one column; all columns should now be marked. Choose FORMAT SECTION in the FORMAT menu. In the opened dialog you can change the gap width to 0%.

7.1.3 Sum Curves

In Tab. 7.2 and in all the following frequency tables we created a column for the “Sum of Relative Frequencies”. The idea behind taking the sum was to obtain the frequency for all entries less than or equal to the corresponding attribute value. For instance, we found that the frequency of grades equal to or better than (= less than or equal to) grade B turned out to be 0.415 = 41.5% (see Tab. 7.2), or that micro or small loans accounted for 64.3% of the total loan portfolio based on Tab. 7.6. In order find out how many numbers in the list of Tab. 7.4 will be between 16 and 40 we could either add the frequencies of the three classes k = 2,


3 and 4 or simply subtract from the frequency sum for class k = 4 (0.85) the frequency sum for class k = 2 (0.0833), giving 0.7667. Of course, in short lists it can also be calculated by adding the frequencies of the classes k = 2, 3 and 4. However, for very long lists it is much more convenient to obtain the frequencies of intervals by simply subtracting the frequency sums.

It is now quite straightforward to represent the frequency sum in a column diagram too. The result for the data of Tab. 7.2 is given in Fig. 7-4. In order to accentuate continuity the columns of the sum frequencies are usually sketched without gaps. In each column you will find the relative frequency of each group on top of the cumulative relative frequencies of the groups left of it. The dotted horizontal lines are added in order to highlight this.

Sum of Relative Frequencies

Grades

Rel

ativ

e Fr

eque

ncie

s

0.00

0.20

0.40

0.60

0.80

1.00

1.20

A B C D F

Figure 7-4: Column diagram of the frequency sum of grades

If we illustrate the frequency sums of Tab. 7.6 in the same way, we obtain a very similar diagram (see Fig. 7-5). Both diagrams have in common that they cumulate towards the value 1, because the sum of the relative frequencies is always 1.


Sum Curve (Loan Portfolio)

Catagories

Rel

ativ

e Fr

eque

ncie

s

0 .0 0 0

0 .2 0 0

0 .4 0 0

0 .6 0 0

0 .8 0 0

1 .0 0 0

1 .2 0 0

M ic r o S m a ll M e d iu m L ar g e

63.4%

Figure 7-5: Frequency sum of loans as a column diagram

We have already pointed out that the sum curve is very informative if we are interested in the relative frequencies of more than one class. For instance, the arrow on top of the second column at the value 0.643 on the vertical axis shows that 64.3% of all loans are either micro or small. However, if we were interested in estimating some cumulative value not exactly on top of one the columns, the diagram would not help us. In this case, however, a logical assumption may help us: If we assume that the individual loans in each class are equally distributed, meaning that each loan size within a class has the same probability, a diagonal line in each column of relative frequencies would represent the distribution within the class. If we now transfer the diagonals together with the columns into the cumulative sum diagram, all pieces of the diagonals are connected to each other. Together they form a so-called polygon of linear pieces of lines whose form is usually s-shaped. This line is called the distribution curve of the underlying statistic.


Sum Curve (Loan Portfolio)

Catagories

Rel

ativ

e Fr

eque

ncie

s

0 .0 0 0

0 .2 0 0

0 .4 0 0

0 .6 0 0

0 .8 0 0

1 .0 0 0

1 .2 0 0

M ic r o S m a ll M e d iu m L ar g e0 1000 8500 10000 20000 30000 90000

Figure 7-6: Cumulative sum curve and distribution function

With reference to Fig. 7-6 we can now estimate – due to the many assumptions the consequences are always only rough estimates and never real results – that 50% of all loans in the portfolio may be smaller than or equal to EUR 8,500. Or we may conclude that about 78% of all loans are less than or equal to EUR 20,000. The difference of the two values, namely 28, tells us that 28% of the loans are consequently between EUR 8,500 and EUR 20,000.

If we made the group sizes smaller, the distribution curve would be smoother. In the limit the curve loses its property of being a linear polygon. Instead it is a smooth s-shaped curve, which for continuous distributions is finally the distribution function.

7.1.4 Base Problem

Statistics deals with large sets of entries. This makes it absolutely necessary to use statistical software. For this reader we apply the broad array of statistical functions available in Excel.

In the classroom we can discuss only small examples and illustrate the topics using statistical lists with only a few entries. In order to provide exercises which are more realistic, we will draw up a list with 500 entries. It will be generated by means of random numbers using Excel.


For most of the following exercises we will formulate a special question related to this problem. Due to the size of the data set, it can only be answered using Excel.

This problem will be called the “Base-Problem”. Each student must generate this example on his/her own computer. None of the problems generated will be the same, but their properties will be statistically identical.

In order to generate the problem, prepare a table with 20 columns and 10 rows. In the upper left corner assign the following Excel function to the first coefficient:

= RANDBETWEEN(0;9)

This generates a random number between 0 and 9. Track the formula over the complete 10×20 matrix, generating 10 random numbers in each column. Place the sum of the ten random numbers in an extra row. We refer to this row as the generating “base line” for the final list. The base line changes its values randomly with every input in Excel.

Now we will set up the problem in a separate table of 25 rows and 20 columns.

• COPY the base line.

• PASTE it to the next empty line of the final list applying the special paste PASTE VALUE.

• Repeat this procedure for the 25 lines of the final table. The easiest way to do this is to use function key F4 repeatedly for each empty row.

The table is now complete. Save it under the name “Base-Problem”. We will refer to this problem under this name regularly throughout this chapter.

The following Tab. 7.7 contains the data generated as the sum of 10 random numbers in the range 0 to 9. It is inserted only to provide an example with which you can compare your own list. The head row and the head column (both bold and italic) are not part of the list. They are added only for better orientation.


ATTENTION

The problem you have generated will contain completely different entries. Do not allow this to confuse you: the derived results (e.g., average, median, variance, standard deviation etc.) may also differ slightly from the given values. However, the differences will be so minor that they are insignificant. All lists generated using the above method have the same statistical properties.

38 35 49 38 41 41 54 43 49 38 32 37 62 48 19

43 40 47 45 51 36 54 28 29 66 50 60 43 55 40

67 36 38 51 37 53 49 48 41 47 52 55 36 25 31

62 46 35 56 46 36 57 49 40 59 42 58 51 63 39

40 51 52 47 36 52 57 53 48 37 33 48 51 47 31

39 38 52 41 54 45 41 26 45 51 55 40 40 44 41

36 36 37 42 55 40 42 52 57 27 40 56 35 53 34

32 57 37 42 54 52 44 57 34 38 38 59 54 45 39

43 30 59 44 30 46 39 42 55 51 53 31 44 38 52

38 29 43 41 54 41 43 43 51 52 43 55 37 52 58

33 63 41 39 49 41 44 54 47 40 60 55 43 37 43

42 37 42 43 37 48 40 34 39 53 37 47 55 29 38

41 38 31 62 60 55 35 45 54 61 37 46 32 29 36

45 46 36 38 47 49 47 45 32 48 39 38 47 46 47

52 44 68 46 43 46 45 47 31 64 42 49 48 45 38

44 54 52 49 54 38 53 57 58 53 46 32 34 46 18

40 40 43 60 26 51 42 37 50 48 40 34 46 39 44

49 35 41 37 34 48 64 46 44 52 33 36 45 41 47

51 36 57 51 54 29 54 55 38 57 47 42 26 43 50

56 49 46 52 56 47 46 39 59 50 26 42 40 45 49

58 45 41 41 49 36 44 52 37 43 49 48 42 48 31

45 65 43 48 61 49 34 53 67 27 41 47 36 54 42

50 54 48 58 55 47 31 55 50 36 61 60 39 71 35

49 40 51 47 51 58 51 50 39 65 51 48 25 43 36


45 70 43 36 42 55 42 42 56 61 40 32 54 27 42

47 35 46 40 49 49 60 46 66 61 42 40 57 47 47

57 40 56 41 39 50 28 46 46 49 38 43 33 43 46

49 24 37 39 51 47 57 47 43 50 40 27 60 44 38

48 28 54 31 40 52 38 47 50 53 52 46 48 38 33

51 40 32 42 44 27 44 40 43 44 52 53 39 50 47

46 40 33 38 52 43 45 40 28 62 49 57 50 44 33

43 52 47 41 27 53 17 47 40 34 48 54 62 46 40

55 52 35 57 44 40 47 41 40 55 49 29 54 32 60

54 45 42 41 58

Table 7.7: “Base-Problem”


EXERCISE 7.1: PRESENTATION TECHNIQUES

1. After reweighing 40 0.5kg packages of sugar Anna found the following values (in grams = g).

492 497 497 491 499 504 495 503 499 504

495 496 502 500 499 500 508 502 500 499

498 501 493 506 494 511 496 488 494 500

506 503 497 497 498 504 499 504 501 498

Set up frequency tables with:

a) Class width = 2g

b) Class width = 4g

2. The following list contains the grades scored by German students in their maths exam. Set up a frequency table and estimate the mean grade.

2 2 4 3 2 1 1 4 2 5 3 4 3 2

1 1 2 4 3 3 3 5 4 2 5 3 1 3

5 1 1 4 3 2 4 3 5 1 2 2 3 4

3 3 4 2 5 1 2 1 1 2 4 4 3 5

3 2 3 2 4 5 2 1 4 3 5 2 3 2

Use Excel – if available – in order to present the distribution as a column diagram.


3. Given the following list of 60 numbers:

25 6 16 30 26 5 20 30 24 37

35 17 19 18 33 26 45 16 30 26

49 22 21 48 26 27 19 29 32 7

12 13 23 19 27 39 29 28 45 37

34 1 24 28 28 33 17 29 38 36

21 25 21 45 20 12 40 48 22 41

a) Set up a list with grouped data. Choose the class sizes (0-10, 10-20, 20-30, etc.) with the right values included, and the left values excluded.

b) Use Excel to set up the table of grouped data.

c) Calculate the sum of absolute frequencies and present them in a column diagram. Close the gaps between the columns in the diagram.

d) Add the s-shaped piecewise linear sum curve.


4. In the last statistics exam the teacher set up the following frequency table of marks:

Marks Frequency from to

0 20 50

21 50 90

51 75 170

76 100 90

Sum 400

a) Sketch the distribution function via relative frequencies, sum of rel. frequencies and the sum curve.

b) Estimate the percentage of students who scored 60 or higher.

5. Use the “Base-Problem”:

a) Guess the average value of the 500 numbers.

b) Define class widths of 10 (11-20, 21-30, etc.) and set up the table of frequencies for grouped data.

c) Create a histogram and estimate the mean value.

d) Choose the alternative class width of 5 and repeat b) and c). Compare the results dependent on the class size.

6. Use the “Base-Problem” and the frequency tables of problem 5:

a) Create a diagram of the sum of relative frequencies and the sum curve.

b) Mark and estimate the percentage of numbers between 37 and 54.

c) Repeat the process for the table with class size 5 and compare the estimates.


ANSWERS 7.1: PRESENTATION TECHNIQUES

1. a) Min is 488 and Max is 511

Class Freq. Class Freq.

488-489 1 500-501 6

490-491 1 502-503 4

492-493 2 504-505 4

494-495 4 506-507 2

496-497 6 508-509 1

498-499 8 510-511 1

Sum 40

b)

Class Freq. Rel. Freq.

488-491 2 5%

492-495 6 15%

496-499 14 35%

500-503 10 25%

504-507 6 15%

508-511 2 5%

Sum 40 100%


2.

Grade Freq. Rel. Freq.

1 12 17.1%

2 18 25.7%

3 18 25.7%

4 13 18.6%

5 9 12.9%

Sum 70 1

0%

5%

10%

15%

20%

25%

30%

1 2 3 4 5

Grade Distribution

Grades

Rel

ativ

e Fr

eque

ncie

s

3.

Groups Middle Abs.Freq. Rel.Freq. Freq.Sum

1-10 5 4 0.0667 0.0667

11-20 15 13 0.2167 0.2833

21-30 25 25 0.4167 0.7000

31-40 35 11 0.1833 0.8833

41-50 45 5 0.0833 0.9667

51-60 55 2 0.0333 1.0000

Sum

60 1


0.0

0.2

0.4

0.6

0.8

1.0

1.2

1-10 11-20 21-30 31-40 41-50 51-60

Sum Curve

Sum

of

Rel

ativ

e Fr

eque

ncie

s

Groups

4. a) and b)

0.0

0.1

0.1

0.2

0.2

0.3

0.3

0.4

0.4

0.5

0-20 21-50 51-75 76-100

Marks in Statistics Exam

Rel

ativ

e Fr

eque

ncie

s

Groups


Groups

Sum

of

Rel

ativ

e Fr

eque

ncie

s

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0-20 21-50 51-75 76-100

Sum Curve of Marks

60

0.53

The estimated percentage of students with marks better than 60 is 100% - 53% = 47%.

5. Class widths of 10:

Group till Freq. Rel. Freq. Freq. Sum

10-20 20 3 0.6% 1%

21-30 30 25 5.0% 6%

31-40 40 134 26.8% 32%

41-50 50 195 39.0% 71%

51-60 60 120 24.0% 95%

61-70 70 22 4.4% 100%

71-80 80 1 0.2% 100%

500 100%


0%

5%

10%

15%

20%

25%

30%

35%

40%

45%

1-10 21-30 31-40 41-50 51-60 61-70 71-80

Groups

Rel

ativ

e F

requ

enci

es

Frequency Distribution (Width: 10)

45

Estimated mean is 45.

Class widths of 5:

Group till Freq. Rel. Freq. Freq. Sum

16-20 20 3 0.6% 0.6%

21-25 25 3 0.6% 1.2%

26-30 30 22 4.4% 5.6%

31-35 35 39 7.8% 13.4%

36-40 40 95 19.0% 32.4%

41-45 45 96 19.2% 51.6%

46-50 50 99 19.8% 71.4%

51-55 55 82 16.4% 87.8%

56-60 60 38 7.6% 95.4%

61-65 65 16 3.2% 98.6%

66-70 70 6 1.2% 99.8%

71-75 75 1 0.2% 100.0%

500 100.0%


0%

5%

10%

15%

20%

25%

16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55 56-60 61-65 66-70 71-75

Rel

ativ

e Fr

eque

ncie

s

Groups

Frequency Distribution (Width: 5)

45

Estimated mean is 45.

6.

0%

20%

40%

60%

80%

100%

120%

1-10 21-30 31-40 41-50 51-60 61-70 71-80

Groups

Sum Curve (width: 10)

Sum

of

rela

tive

fre

quen

cies

Estimated percentage of numbers between 37 and 54 is:

81% – 22% = 59%


Groups

Sum Curve (width: 5)

0%

20%

40%

60%

80%

100%

120%

16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55 56-60 61-65 66-70 71-75Sum

of

rela

tive

freq

uenc

ies

Estimated percentage of numbers between 37 and 54 is:

82% – 20% = 62%


7.1.5 Progress Test “Presentation Techniques”


At the end of the chapter you will find the solutions to the problems. Each solution is reduced to the final answer; it may even be only a number, a symbol, a table or a graph. You should check your answers. If they are correct, you may continue and start the next chapter. Otherwise (if you got the wrong answer or no answer at all) you will be given advice on what to do. In most cases you will be directed to the section in the reader which you should repeat in order to close the gap.

1. Evaluate the sum and simplify the expression:

2

1

2( )n

jj

x a=

− =∑

2. The following table shows the application data for six ProCredit banks and the respective percentages of admissions:

Men Women Bank Applicants Admissions Applicants Admissions

A 224 43% 345 62%

B 267 54% 336 57%

C 127 23% 287 34%

D 117 37% 256 29%

E 191 44% 312 59%

F 263 67% 211 43%

Calculate the weighted average of admissions for men and women.


3. The following table shows the number of disbursed loans per loan officer.

22 17 19 21 22 26 23 21 18 25

29 33 21 26 28 22 31 33 35 27

24 22 23 24 29 30 28 20 31 24

25 26 37 27 34 23 24 36 23 33

a) Set up a tally list of frequencies.

b) Draw a column diagram of the relative frequency distribution.

4. Given the following list of 40 numbers.

22 17 19 21 22 26 23 21 18 25

29 33 21 26 28 22 31 33 35 27

24 22 23 24 29 30 28 20 31 24

25 26 37 27 34 23 24 36 23 33

Set up a tally list after grouping the data in steps of 3 (17-19, 20-22, etc.) and draw a column diagram of the relative frequencies, the frequency sum, and the sum curve.

5. Using the sum curve from problem 4, estimate the numbers between 24 and 30. How many numbers really were within this range?

7 . 2 K e y F i g u r e s : C e n t r e


Prerequisites: To work throughfamiliar withdiscussed in calculating with large sets of numbers to be able to usewithout writing long expressions.

Again, we would like to remind the reader that provides a wide range of functions which can be used to calculate the figures needed. In most cases we will refer to the


7.0 Sum Symbol


Tables

Diagrams

Sum Curves


7 . 2 K e y F i g u r e s : C e n t r e P a g e | 187

To work through this section it is necessary to be familiar with the sum symbol introduced and discussed in Section 7.0.1. Very obviously, when calculating with large sets of numbers it is necessary to be able to use notation which allows us to argue without writing long expressions.

Again, we would like to remind the reader that Excel provides a wide range of functions which can be used to calculate the figures needed. In most cases

refer to these functions explicitly.


7.0 Sum Symbol


Mean

Median

Mode


Variance

Standard Deviation

Normal Curve



Having summarised the data by means of graphical tools, we now introduce key figures which also offer certain concentrated information about the statistical list and its attributes.

The figures will be calculated. We differentiate between the various types of information about the centre of the data. The catchwords are mean, median, and mode.

By combining information about the centre of the data with information about their concentration or spread, which will be discussed in the next section, we can obtain quite an accurate picture of the distribution of data without drawing a diagram.

An appropriate and convincing method of analysing distributions of large sets of data with cardinal attribute values is to represent them graphically. This kind of visual analysis is especially useful in presentations.

However, the analysis of diagrams and curves is less effective when we need concentrated information about data for further quantitative analysis. For this purpose it is much more efficient to characterise distributions by key figures = measurement figures. They are constructed in such a way that they offer a comprehensive picture of certain central properties of the distribution. In this section we will calculate figures which characterise certain accentuated positions of the distribution. They can be referred to as the “big M’s”:

• Mean

• Median

• Mode

• M inimum

• Maximum

These figures already allow us to gain a very good understanding of the distribution.


7.2.1 Mean

The most prominent figure for describing distributions is the statistical mean (often also called: average, midpoint, average value, mean value, or arithmetic mean). The many synonyms are a strong indicator that almost everybody has a rough idea of how to calculate a mean, because it is quite common in daily life to argue with averages. We calculate the average maturities of loans, the average days of arrears, the average fuel consumption of our car, the average speed of the finishers in a marathon, the average family income, or the average tax.

How do we calculate these average values? Most of us are quite familiar with the process of adding up all the attribute values of which we want to take the average, and dividing the sum by their number.

EXAMPLE

We assume that the maturities of five loans are: 1 year, 6, 9, 8 and 6 months, respectively. What is the average maturity of the five loans?

In order to calculate the average maturity, we add up the attribute values.

ATTENTION

When adding up attribute values, make sure that they all have the same dimension. In other words, do not add apples and pears if you want to know the number of apples in the basket.

That means, before adding together the maturities, they all have to be transformed either into years or into months.

Using the dimension “months”, we get the sum:

12 6 10 8 6 42m= + + + + = months

Dividing this by the number of loans gives: 42

8.45

m= = months

The average maturity of the five loans is: 8.4 months.


In the following discussion we refer to a statistical list consisting of a set of entries having (cardinal) attribute values of the same dimension.

STATISTICAL L IST

If we have n observations 1 2, , , nx x x… we call { }1 2, , , nL x x x= … a set or

list of observations.

n is the number of entries in the list.

The sum of all observations or attribute values is:

1 21

...n

n jj

x x x x=

+ + + =∑

The sum symbol, introduced in Section 7.1.1 (see: Page 147), will now be used extensively. If we divide the sum by the number of entries in the list we will get the average value of all attribute values. We refer to it as the statistical mean of the attribute values, or – even more precisely – as the arithmetic mean of the values.

ARITHMETIC MEAN

The arithmetic mean (for short: mean or average) of n observations ix

from the list { }1 2, , , nL x x x= … is:

1

1

1

n

j nj

jj

x

x xn n

=

== =∑

∑

The mean is commonly denoted by the un-indexed symbol topped by a bar (x ).

EXCEL : In Excel the function AVERAGE returns the average or arithmetic mean of its arguments.


EXAMPLE

Use Excel to calculate the average of the numbers in Tab. 7.4 (see Section 7.1.2; Page 163). The following Fig. 7-7 is a screenshot of the table using MS Office 2007. The calculated value of the average is 29.1666667.

Figure 7-7: Statistical figures in the Excel status bar

EXCEL : Excel provides some additional statistical features which are very useful for quick information: If you mark any array in Excel, Office 2007 automatically calculates different statistical figures for the marked numbers and displays them in the status bar at the bottom of the screen (see screenshot in Fig. 7-7):

Average: 29.16667 = the arithmetic mean of the numerical attributes.

Count: 60 = the number of entries in the list.

Numerical Count: 60 = the number of numerical entries in the list. If the list of attributes contains numerical and non-numerical values, the other results are calculated on


the basis of the numerical values only.

Min: 1 = the minimum numerical value in the list.

Max: 66 = the maximum numerical value in the list.

Sum: 1750 = the sum of all numerical attributes.

If one or more attribute values appear more than once in the list, the calculation can be simplified. We assume that the attribute kx appears in

the list with frequency kf . The list of entries now consists of the m

different attributes and the list of corresponding frequencies:

1 2{ , ,..., }mL x x x= and 1 2{ , ,...., }mF f f f=

The total number of entries is in this case:

1

m

kk

n f=

=∑

Every attribute value appears in the sum of attributes as often as the frequency indicates. Hence, the arithmetic mean is now:

1

1

1

1

m

k k mk

k kmk

kk

x f

x x fn

f

=

=

=

⋅= = ⋅∑

∑∑

In this formula we can multiply the sum by the constant factor 1n to get:


1

1 1

1

1

m

k k m mk k

k k kmk k

kk

x ff

x x f xn n

f

=

= =

=

⋅= = ⋅ = ⋅∑

∑ ∑∑

We defined the absolute frequency divided by the total number of entries

as the relative frequency kk

fh

n= (see: Section 7.1.2; Page 167). Thus,

we arrive at an alternative formula for the arithmetic mean:

1 1

1 m m

k k k kk k

x x f x hn = =

= ⋅ = ⋅∑ ∑

EXAMPLE

In Germany exams are graded using a number system. Grade 1 is the best grade (“excellent”), while grade 5, the lowest, means “failed”. The following table contains all grades for the 16 students in the statistics course.

Student i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Grade jg 2 2 4 3 2 1 1 4 2 5 3 4 3 2 1 3

Counting the frequencies and the relative frequencies of the grades we obtain:

Grades k 1 2 3 4 5 Sum

Frequency kf 3 5 4 3 1 16

Rel. Frequency kh 0.1875 0.3125 0.25 0.1875 0.0625

Now we calculate the average grade using the three different formulas.

• Using the grades only:


1 1 2116 16 8

1

1(2 2 4 ... 1 3) 42 2.625

n

jj

g gn =

= = ⋅ + + + + + = ⋅ = =∑

• Using the frequencies:

1 1 4216 16

1

(1 3 2 5 3 4 4 3 5 1) 2.625m

k knk

g g f=

= ⋅ = ⋅ ⋅ + ⋅ + ⋅ + ⋅ + ⋅ = =∑

• Using the relative frequencies:

1

1 0.1875 2 0.3125 3 0.25 4 0.1875 5 0.0625m

k kk

g g h=

= ⋅ = ⋅ + ⋅ + ⋅ + ⋅ + ⋅∑

1

2.625m

k kk

g g h=

= ⋅ =∑

We can see that the results are exactly identical, regardless of which formula we use.

EXCEL : In addition to the Excel function AVERAGE it is sometimes very helpful to apply the function SUMPRODUCT. This returns the sum of the product of the coefficients of two congruent arrays. Congruent means the arrays must be of identical size. Identically positioned coefficients of the two arrays are multiplied and added. For instance, we can apply this Excel function when

calculating 1

m

k kk

g f=

⋅∑ or 1

m

k kk

g h=

⋅∑


ARITHMETIC MEAN IN THE CASE OF GROUPED DATA

If the entries and their frequencies are exactly enumerated, as was the case in the above example, all three formulas yield identical exact results. However, if the entries are grouped in classes with the middle point as the reference value, the results differ slightly.

Class Left Right Middle Abs.Freq. Rel.Freq. Freq.Sum k kl kr km kf kh

1 0 15 7.5 5 0.0833 0.0833

2 15 25 20 16 0.2667 0.3500

3 25 30 27.5 23 0.3833 0.7333

4 30 40 35 7 0.1167 0.8500

5 40 50 45 5 0.0833 0.9333

6 50 70 60 4 0,0667 1,0000

Sum 60 1

Table 7.8: Grouped frequencies (Copy of Tab. 7.5)

For example, let us look again at the data of Tab. 7.4 which is also presented in the screenshot in Fig. 7-7. The analysis is given in Tab. 7.8 which is a copy of Tab. 7.5. The calculated average of the 60 entries jx

is: 29.1667x =

We can apply the formula with the frequencies only if we have one representative value for each group. A value which offers itself naturally could be the middle point of the interval (group), which is also calculated in Tab. 7.8. If we apply the frequency-related formula using the middle points as representative values we obtain:

1700160 60

(7.5 5 20 16 27.5 23 35 7 45 5 60 4) 28.3333x = ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ = =

Obviously, there is a difference between the two results which derives from the inaccuracy of the representative value. The middle point of the interval is only in exceptional situations the true average of the entries in the group.


This might be especially true in the case of loans. We defined for instance the group of micro loans as consisting of loans between 0 and 4000. In this group there will not be any entries 1, 2, or even 99. The majority of all micro loans are probably close to or exactly equal to 4000. Therefore, 2000 is never the representative value of the entries in this group. It will be much closer to the right-hand interval bound. This is probably also the case in the other groups; and if so, the arithmetic means calculated on the basis of the middle points of each interval underestimate the true average.

Applying the formula with the sum-product of middle point times relative frequencies gives the same result as the last line:

6

1

28.3333k kk

x m h=

= ⋅ =∑

7.2.2 Median

Let us have a look at the following list of 60 numbers in Tab. 7.9:

List of Entries

2 7 2 9 3 4 1 4 9 4 3 8

1 9 8 0 3 4 0 2 3 9 4 0

3 1 4 7 9 2 0 4 1 0 2 4

1 0 2 2 9 3 3 8 9 9 8 9

3 5 1 0 9 3 9 2 4 1 8 8

Table 7.9: List of 60 numbers

The arithmetic mean of these 60 numbers is 4.2. It is probably natural to guess that 50% of the entries in the last list were above average, and 50% were below. However, this guess is not correct. In fact, only 33.3% of the entries were above, and about 6.67% were below average.


To understand this apparent contradiction better, let us start with a hypothetical list of data:

L = {1, 9, 1, 9, 1, 1, 1, 9, 1, 1}

The average of the entries is 3410

3.4= ; i.e. 7 entries in the list are below

and only 3 entries are above average. If we arrange the entries in ascending order and divide them into two parts

L = {1, 1, 1, 1, 1 | 1, 1, 9, 9, 9}

with the same number of entries, the divider lies between two entries “1”. The value where the entries are divided by size into two equal halves is called the median. In the example the median is therefore 1, the average is 3.4.

0%

25%

50%

0 1 2 3 4

(a) List: 1, 2, 2, 3; Average 2

0%

25%

50%

0 1 2 3 4

0%

25%

50%

0 1 2 3 4

5 6

5 6 7 8

2,5

Median = Average

(b) List: 1, 2, 2, 5; Average 2.5

(c) List: 1, 2, 2, 7; Average 3

Median

Median

Average

Average

Figure 7-8: Histogram balanced at the average


The histograms in Fig.7-8 explain what is happening. Histogram (a) is derived from the ordered list {1,2,2,3}aL = ; it is symmetric around the

average (of 2), and exactly one half of the area lies on either side of the dotted line.

What happens when the value 3 in the list is increased, say to 5 (the list will become {1,2,2,5}bL = ) or to 7 (with the list {1,2,2,7}cL = )? The

average moves off-centre to the right if one of the numbers is increased in value.

If the histogram is made of wooden blocks attached to a stiff board, the board will always be in balance when supported at the average.

However, although the average moves to the right if the right weight is moved to the right, the area is still divided into two parts with two blocks on either side at line 2. The value dividing the area into two equal parts is called the median. It is another statistical number used to summarise data.

MEDIAN

The median medx is the value with half of the entries of a list to the left =

less or equal, and the other half to the right = greater or equal. Since we divide the list of entries by size, we argue with a sorted list where the entries are sorted in ascending order.

If the number of entriesn is odd, medx is exactly the middle element and

therefore an element of the list. In a list with an even number of entries the median lies between two entries in the middle of the list. In this case the median is the average of the two middle entries. If they are equal, the median is this value. If they are different, the median is the arithmetic mean of the two adjacent numbers. In this case the median is not an entry of the list.

EXCEL : In the division “statistical functions” Excel provides the function MEDIAN which returns the median, or the


number in the middle of the set of given numbers. The operand of this function can be any array.

EXAMPLES

1. Let 1L be a list of random numbers jx with average 2.95x = :

jx 7 3 2 3 5 2 1 3 0 5 1 6 2 1 4 9 1 0 2 1 4

In order to illustrate the median we arrange the numbers in ascending order:

jx 0 0 1 1 1 1 1 2 2 2 2 3 3 3 4 4 5 5 6 7 9

The list contains an odd number (21) of entries; hence, the median is the element in the middle of the list, which is the shaded number:

2medx =

In this example the median is smaller than the average: Therefore, the smaller entries in the distribution “outweigh” the larger ones. The following frequency distribution of the numbers in Fig. 7-9 shows the left-sided concentration of the numbers. This type of distribution is called a right-tail distribution.


Freq

uenc

ies

0

1

2

3

4

5

6

1 2 3 4 5 6 7 8 9 10

Numbers

Frequency Distribution

Figure 7-9: Right-tail distribution

Now look at the list 2L with 20 entries; their average is 6.15x = .

jx 7 6 2 3 5 6 8 7 8 5 8 6 8 7 4 9 8 9 0 7

Again, we arrange the numbers in the list in ascending order:

jx 0 2 3 4 5 5 6 6 6 7 7 7 7 8 8 8 8 8 9 9

The first half with the smaller values is shaded. Since we have an even number of entries the median lies between 7 and 7, the mean of which is of course also 7. Therefore, the median of the above list is:

7medx =

Now the median is larger than the average the distribution. The larger numbers in the list outnumber the smaller ones. The frequency distribution of the entries in Fig. 7-10 illustrates that clearly. This is called a left-tail distribution.


Numbers

0

1

2

3

4

5

6

0 1 2 3 4 5 6 7 8 9

Freq

uenc

ies


Figure 7-10: Left-tail distribution

In general, the average is greater than the median in right-tail distributions, the average is less than the median in left-tail distributions, and the average equals the median in symmetric distributions (see Fig. 7-11).

Average

Median

Average

Median

50%

Average = Median

Left tail: Average < Median Right tail: Average > MedianSymmetric: Average = Median

50% 50%

Figure 7-11: The tails of frequency distributions

Taking the average is a powerful way of summarising data, but it also hides the diversity of the data. However, average and median together provide some additional information about the shape of the frequency distribution.


7.2.3 Mode

Knowing the most frequent entry in the list can also provide additional information about the statistical properties of the attribute values. For instance, if we want a “representative value” for the micro loan portfolio, it may make more sense to choose the most frequent value of all micro loans, rather than the middle point of the micro loan group. (Remember that we discussed the problem of selecting the representative value when we calculated the statistical mean and encountered a bias due to the selection.)

Hence, it is quite common in statistics to look at

• the minimum value,

• the maximum value, and

• the most frequent value in the list.

M INIMUM VALUE

The minimum value is simply the entry mx such that all other entries in

the list are greater than or equal to the minimum: j mx x≥ for all j.

Accordingly, we will define the maximum value in the list as follows:

MAXIMUM VALUE

The maximum value is simply the entry Mx such that all other entries in

the list are less than or equal to the maximum: j Mx x≤ for all j.

As an example, let us look again at the list of numbers in Tab. 7.4 (see Section 7.1.2; Page 163).


25 15 16 30 26 5 20 30 24 37 21 25 25 17 19 25 28 26 52 26 30 26 21 45 49 22 21 43 26 27 19 29 32 27 30 12 66 13 23 29 27 39 29 28 35 27 40 48 57 1 24 28 28 33 27 29 38 26 63 41

After a quick inspection you will probably find:

• The minimum value is 1mx =

• The maximum value is 66Mx =

It should also be noted that analysing the whole list, although it is relatively short, is not a trivial task any more. For lists with many more entries it is useful to get the information using Excel.

EXCEL : One may use the two functions MIN and MAX, which return the minimum and the maximum values, respectively, of an array.

In addition, watch the status bar at the bottom of your Excel sheet when you mark an array. The minimum and maximum values are among the information displayed.

MODE

The mode modx of a statistical list is defined as the most frequent attribute

value in the list.

In order to find the most frequent value it is useful to look at the frequency distribution. Without this information it would be quite tedious to find this value.

The mode in the list of numbers in the above table is:

modx = 26


EXCEL : The statistical function MODE returns the most frequently occurring entry (attribute value) in an array or range of data.

If more than one value are candidates, i.e. if they appear equally often with maximal frequency, Excel returns only one value. It is the value which appears first of all equally qualified values in the list.

EXAMPLES

1. The mode of the list1 {1,2,4,5,4,2,3,2,1,5}L = is:

2. The list has more than one

most frequent element. Each of the numbers 5, 6 and 7 appear three times. Hence they all are qualified. Excel, though, returns 7 because, of the three candidates, the 7 appears first in the list.

ATTENTION – TYPICAL M ISTAKES

Calculating the average

� minimum + maximum

average 2

≠

� minimum + maximum

average number of entries

≠

mod 2x =

{ }2 2,7,5,6,5, 4,5,6,7,6, 4, 2,7,3L =


EXERCISE 7.2: KEY FIGURES: CENTRE

1. a) Calculate the average of the numbers in the list:

1 -2 4 5 6 -3 0 3 7 -4

3 5 -6 -3 4 5 -3 4 8 5

b) Sort the numbers in ascending order and determine the median.

2. Given is the list of grades according to the German system (from 1 to 5) in the statistics exam:

Student i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Grade 1 3 4 3 2 1 2 4 2 5 3 4 3 2 1 5 2 3 4 3

a) Calculate the average grade.

b) Compile a frequency table.

c) Verify the average by applying the two alternative formulas using the absolute and relative frequencies.


3. The following list of 70 grades was already used in exercise 7.1, problem 2:

2 2 4 3 2 1 1 4 2 5 3 4 3 2

1 1 2 4 3 3 3 5 4 2 5 3 1 3

5 1 1 4 3 2 4 3 5 1 2 2 3 4

3 3 4 2 5 1 2 1 1 2 4 4 3 5

3 2 3 2 4 5 2 1 4 3 5 2 3 2

Use Excel to determine

a) the average,

b) the median,

c) the mode of the list.

4. Use the frequency list which you compiled in exercise 7.1, problem 2, and determine the average without using Excel.

5. Given is a list of 50 randomly generated numbers between 1 and 50:

25 29 16 30 26 5 20 30 24 37

35 17 24 18 33 26 45 16 30 26

21 22 48 28 50 23 19 27 32 7

12 13 7 19 27 39 29 3 45 37

34 1 24 8 28 33 17 29 45 36

a) Group the data in five classes, each with width 10, i.e. 1-10, 11-20, etc., and use a tally list to compile a frequency table.


b) Calculate the average using the absolute and relative frequencies, respectively, and the class middle points (5, 15, etc.).

c) Calculate the exact arithmetic mean of the numbers with Excel, and compare it with the results of 5 b).

6. Given is a list of 14 integers:

-3 4 2 -2 5 -1 1 6 4 1 6 0 1 4

a) Calculate the average, median, and mode.

b) Determine whether the frequency distribution is basically left-tail or right-tail?

7. Use the “Base-Problem”:

a) Calculate the mean, median and mode with the appropriate Excel functions. What conclusions can be drawn from the values of average and median?

b) Use the frequency tables (with class width 10) compiled in exercise 7.1, problem 5 b) and calculate the average again. Compare the result with that of 7 a).

c) Calculate the average using the smaller class width of 5 (see exercise 7.1, problem 5 d)) and compare the result with that of 7 b).


ANSWERS 7.2: KEY FIGURES: CENTRE

1. a)

b)

2. Given is the list of grades according to the German system (from 1 to 5) in the statistics exam:

a)

b) Frequency table.

Grades Freq. rel. Freq.

1 3 15%

2 5 25%

3 6 30%

4 4 20%

5 2 10%

20 100%

c)

3. a) 2.842857

b) 3

c) 2

1.95x =

3.5medx =

572.85

20g = =

2.85g =


4.

5. a)

Right Middle Freq. rel. Freq.

10 5 6 12%

20 15 10 20%

30 25 20 40%

40 35 9 18%

50 45 5 10%

Sum 50 100%

b)

c) The exact mean calculated using Excel is 25.5

6. a) ; ;

b) Right-tail

7. a) ; ; → symmetrical

b)

c)

170 199 2.84286g = ⋅ =

24.4x =

2x = 1.5medx = mod 4x =

45.036x = 45medx = mod 40x =

44.98x =12

1

1

2255545.11

500k knk

x m f=

= ⋅ = =∑


7.2.4 Progress Test “Key Figures: Centre”


At the end of the chapter you will find the solutions to the problems. Each solution is reduced to the final answer; it may even be only a number, a symbol, a table or a graph. You should check your answers. If they are correct, you may continue and start the next chapter. Otherwise (if you got the wrong answer or no answer at all) you should repeat the corresponding units in order to close the gap.

1. Calculate the arithmetic mean, the median, and the mode of the following list of numbers:

{0,7,3, 1, 1,4, 2,0,4,2.5, 4,3.5,1,4,1}L = − − −

2. The average height of ten people in a room is 175 cm. An 11th person who is 180 cm tall enters the room. Find the average height of all 11 people.

3. The average height of 21 people in a room is 175 cm. A 22nd person enters the room. How tall would he have to be to raise the average height of all 22 people by 2 cm?

4. The average of the 40 numbers in exercise 7.1, problem 4 is 26.05. The table of grouped frequencies was:


Group Right Middle Freq. Rel. Freq.

Sum of rel. Fr.

17-19 19 18 3 7.5% 7.5%

20-22 22 21 8 20.0% 27.5%

23-25 25 24 10 25.0% 52.5%

26-28 28 27 7 17.5% 70.0%

29-31 31 30 5 12.5% 82.5%

32-34 34 33 4 10.0% 92.5%

35-37 37 36 3 7.5% 100.0%

Sum 40

On the basis of the grouped data with the middle point as reference value, calculate the average.

5. Given is the sum curve of a distribution of numbers with mean = 25.325.

Sum Curve

Groups

Sum

of

rela

tive

freq

uenc

ies

0.0%

20.0%

40.0%

60.0%

80.0%

100.0%

120.0%

17-19 20-22 23-25 26-28 29-31 32-34 35-37

Mark the median in the graph and decide whether the distribution is left-tail or right-tail.

212 | P a g e


Prerequisites: In this section it is again important to with the sum symbol introduced and discussed in Section 7.0.1. We are still dealing with large amounts of dataallows us to argue without writing long expressions.

Again, we would like to remind the reader that provides a wide range of functions which can be used to calculate the figures needed. In most cases we will refer to t


7.0 Sum Symbol


Tables

Diagrams

Sum Curves


7 . B a s i c s o f S t a t i s t i c s

In this section it is again important to be familiar the sum symbol introduced and discussed in

7.0.1. We are still dealing with large amounts of data. This requires a notation which allows us to argue without writing long expressions.

Again, we would like to remind the reader that Excel provides a wide range of functions which can be used to calculate the figures needed. In most cases we will refer to these functions explicitly.


7.0 Sum Symbol


Mean

Median

Mode


Variance

Standard Deviation

Normal Curve

7 . 3 K e y F i g u r e s : S p r e a d P a g e | 213


Having summarised the data with respect to centre information we now turn our attention to the question of how widely or narrowly they are spread around this centre. It makes a big difference whether the data are mostly concentrated around the average, or are equally distributed over the whole range of data. Sets of data with the same statistical mean can have very different spreads.

While the average was a key figure which many people are very familiar with, the spread figures need some more explanation based on statistical argumentation.

Variance and standard deviation are the most commonly used spread figures. Together with the information about the concentration of the data they provide quite an accurate picture of the distribution of data without our having to draw a diagram.

The average of the following list 1L of 15 numbers is.

1L : jx 2 3 3 3 4 1 5 1 3 2 3 2 3 6 4

We get the same average 3x = for the second list 2L .

2L : jx 2 7 4 -2 4 1 3 3 -2 4 5 6 -1 3 8

However, if we look at the frequency distribution of the two lists, we will notice a distinctive difference between them. The attribute values of list

1L are concentrated narrowly around the average of 3x = while the

values of list 2L are widely spread.


Freq

uenc

ies

Distribution in List 1

0

1

2

3

4

5

6

7

1 2 3 4 5 6

Numbers

Figure 7-12: Narrowly spread attribute values

Numbers

0

0.5

1

1.5

2

2.5

3

3.5

-2 -1 0 1 2 3 4 5 6 7 8

Distribution in List 2

Freq

uenc

ies

Figure 7-13: Widely spread attribute values


In order to quantify the difference between the two statistical lists we obviously have to characterise the distances between the values. Since distances need to be based on a metric, the following calculations of spread figures can only be applied if the attribute values are cardinal, i.e. measured by numbers. In this case we can determine variance and standard deviation.

7.3.1 Variance

When we talk about a distance we immediately think of the range or interval between two points, one of them being the reference point to which we measure the distance. A “natural” reference point could be the origin (zero point) of the array of real numbers. However, it does not make sense to refer all kinds of numbers, which might range up to 100,000 (e.g., number of outstanding loans), to a point which is not associated with the statistical list itself. A much better choice is therefore the arithmetic mean.

A first idea to define “spread” might be to calculate the distances from each attribute value of the list to the mean point and calculate the arithmetic mean of those distances.

Carrying out this operation for the above list 1L of 15 cardinal numbers

with the arithmetic mean 3x = we get:

1L : jx 2 3 3 3 4 1 5 1 3 2 3 2 3 6 4

( )jx x− -1 0 0 0 1 -2 2 -2 0 -1 0 -1 0 3 1

Table 7.10: Number list

The average distance is the sum of all distances divided by the number of entries. The sum of the above distances is equal to zero. Therefore the mean is also zero. Interpreting this result would mean: There is no measurable spread of the numbers around the mean value. “No spread” would imply that all numbers are equal to the mean. This is obviously not


the case; therefore the implication is false, which means that the method of measuring spread is not appropriate.

What caused the failure?

Apparently, we produce “negative distances” when subtracting the mean from numbers which are smaller. But negative distances do not make sense when discussing spread, and – this caused the zero mean – the negative figures cancel out the positive ones. The question is therefore how to handle the fact that negatives cancel out the positives? A measure for the spread of the numbers is naturally the distance between them. In order to overcome the sign, one can calculate the squares of the distances.

1L : jx 2 3 3 3 4 1 5 1 3 2 3 2 3 6 4 sum

( )jx x− -1 0 0 0 1 -2 2 -2 0 -1 0 -1 0 3 1 0

2( )jx x−

1 0 0 0 1 4 4 4 0 1 0 1 0 9 1 26

Table 7.11: Calculating the variance

The sum of the squares does not make sense either, because it depends basically on the number of entries in the list. Therefore, we calculate the mean of the squares. Following the concept of the arithmetic mean, we have to divide the sum by the length of the list, i.e. the number of entries in the list. The resultant figure is called variance.

Thus, the variance of list 1L is:

2615

1.7333=


VARIANCE

The variance of a statistical list L of n numbers jx is defined as the mean

of the attribute values’ squared distances from the mean:

2

1 2

1

( )1

( )

n

j nj

jj

x x

VAR x xn n

=

=

−

= = −∑

∑

EXAMPLE

Let us calculate the variance of the numbers in list 2L with the same

arithmetic mean of 3x = as list 1L :

2L : jx 2 7 4 -2 4 1 3 3 -2 4 5 6 -1 3 8 ∑

( )jx x− -1 4 1 -5 1 -2 0 0 -5 1 2 3 -4 0 5 0

2( )jx x−

1 16 1 25 1 4 0 0 25 1 4 9 16 0 25 128

Dividing the sum of the squares by 15 gives: 12815

8.5333VAR= =

If we now compare the variances of the two lists 1L and 2L we realise

that there is a significant difference in the margin of the result. The variance of list 1L with numbers narrowly spread around its mean is

1( ) 1.7333Var L = which is relatively small compared to the variance of

list 2L with 2( ) 8.5333Var L = .

EXCEL : If you apply the function VAR from the statistical toolbox the results differ surprisingly from the calculated values above. However, if you look at the definition of the


function VAR you will find: “Estimates variance based on a sample (ignores logical values and text in the sample)”.

This explains the difference. Statisticians distinguish between the variance for the whole population – which is calculated by the above formulas – and a slightly different formula, which should be used for samples. The latter is called the unbiased variance. The two formulas differ in the denominator by 1.

The unbiased variance is:

2

1 2

1

( )1

( )1 1

n

j nj

u jj

x x

VAR x xn n

=

=

−

= = −− −

∑∑

In Excel the function VAR returns the unbiased variance. For our purpose – we are almost exclusively using statistics for whole sets of data, i.e. for populations rather than samples – we need to apply the function VARP, where the P stands for “population”. The definition of this function differs from the former quite significantly: “Calculates variance based on the entire population…” Hence, if we apply the function VARP the calculated variances are exactly those that we calculated for lists 1L

and 2L .

AN ALTERNATIVE FORMULA FOR THE VARIANCE

A few algebraic calculations involving the sum symbol can significantly simplify the formula for the variance and especially its application.

2 2 2

1 1

1 1( ) ( 2 )

n n

j j jj j

VAR x x x x x xn n= =

= − = − ⋅ +∑ ∑ ← after squaring

2 2

1 1 1

1 1 12

n n n

j jj j j

VAR x x x xn n n= = =

= − ⋅ +∑ ∑ ∑ ← each term is a sum


2 2

1 1 1

1 1 12 1

n n n

j jj j j

VAR x x x xn n n= = =

= − ⋅ + ⋅

∑ ∑ ∑ ← The terms 2x and

2x are independent of the sum index j and can therefore be taken out of the sum and placed as factors before it.

2 2

1

1 12

n

jj

VAR x x x x nn n=

= − ⋅ + ⋅ ⋅∑ ← by replacing: 1

1 n

jj

x xn =

=∑ (in the

round brackets) and 1

1n

j

n=

=∑ [in the square brackets]

2 2 2 2 2

1 1

1 12 1

n n

j jj j

VAR x x x x xn n= =

= − + ⋅ = −∑ ∑ ← after composing equal

terms

Finally we get:

2 2 2

1 1

1 1( )

n n

j jj j

VAR x x x xn n= =

= − = −∑ ∑

ATTENTION

The above algebraic exploitation is a nice example of the application of the sum symbol and the necessity of brackets (see: the beginning of this chapter; Page 147) in order to keep the uniqueness of the expression:

The sum symbol replaces a pair of brackets: 3

1 2 31

jj

x x x x=

= + +∑ ;

therefore:

( )3

1 2 3 1 2 31

jj

c x x x x c x x x=

⋅ = + + = ⋅ + +∑

If the operand of the sum is a sum itself, brackets must be used, in order to avoid any ambiguity:


( ) ( ) ( ) ( )3

1 2 31

jj

x a x a x a x a=

+ = + + + + +∑

Hence, without brackets the term b is not the operand of the sum symbol and therefore a single summand:

( )3

1 2 31

jj

x a x x x a=

+ = + + +∑

The brackets on the right-hand side may be deleted. They are used only for demonstration.

VARIANCE RELOADED

An alternative way of calculating the variance is: 2 2

1

1 n

jj

VAR x xn =

= −∑

Notice that the first term on the right-hand side is the mean of the

squared values 2jx and that the last term 2x is not part of the summation

denoted by the sum symbol. Therefore, it is helpful to remember that:

“T HE VARIANCE IS THE MEAN OF THE SQUARES MINUS THE SQUARE OF

THE MEAN .”

Calculating the variance “by hand”, i.e. if Excel is not available, is much more efficient if you apply the above formula.


EXAMPLE

Calculate the variance of the numbers in list 2L with the arithmetic

mean of 3x = using the above formula:

2L : jx 2 7 4 -2 4 1 3 3 -2 4 5 6 -1 3 8 sum

2jx 4 49 16 4 16 1 9 9 4 16 25 36 1 9 64 263

The mean of the squares is 26315

17.5333= ; thus:

17.5333 9 8.5333VAR= − =

VARIANCE WITH FREQUENCY TABLES

Frequency tables (see Section 7.1.1; Page 159) are a first step towards concentrating large sets of data. We took advantage of this form when calculating the arithmetic mean. Since the variance is simply another mean, we can in the same manner modify the above formula using the frequencies:

2

1

1

( )m

k kk

m

kk

f x x

VAR

f

=

=

⋅ −=∑

∑

Since 1

m

kk

f n=

=∑ we obtain:

2

1

1( )

m

k kk

VAR f x xn =

= ⋅ −∑


We can also multiply the factor 1n into the sum and replace the absolute

frequencies using:

kk

fh

n=

by the relative frequencies, and obtain the third alternative formula:

2

1

( )m

k kk

VAR h x x=

= ⋅ −∑

The formula “variance = mean of the squares minus square of the mean” is similarly applicable if the statistical list includes frequencies:

• Using the absolute frequencies: 2 2

1

1 m

k kk

VAR f x xn =

= ⋅ −∑

• Using the relative frequencies: 2 2

1

m

k kk

VAR h x x=

= ⋅ −∑

Remember, applying these formulas reduces the calculation quite substantially.

EXAMPLE

The statistical list 1L with average 3x = can be replaced by the

frequency table. Applying the different formulas we always obtain the same results:


entry 1 2 3 4 5 6 sum frequency 2 3 6 2 1 1 15 relative freq. 0.133 0.2 0.4 0.133 0.067 0.067 1

2( )jx x− 4 1 0 1 4 9 2( )k jf x x⋅ −

8 3 0 2 4 9 26 2( )k jh x x⋅ −

0.533 0.2 0 0.133 0.267 0.6 1.733 2

jx 1 4 9 16 25 36 91 2

k jf x⋅

2 12 54 32 25 36 161 2

k jh x⋅

0.133 0.8 3.6 2.133 1.667 2.4 10.73

The results are the same in all cases. The originally calculated variance 1.7333 will also be obtained if the sum in the line with the absolute frequencies (26) is divided by the sum of frequencies (15): The same figure is also obtained as the sum in the line with the relative frequencies.

Alternatively, we can obtain the variance by dividing the sum in the line with the absolute frequencies (161) by the sum of frequencies

(15) giving 1.7333 after subtracting the square of the mean ( 2 9x = ), which is also obtained when subtracting the squared mean from the sum in the line with the relative frequencies (10.73).

VARIANCE FOR GROUPED DATA

In Section 7.2.1 (see: Page 190) we discussed how the arithmetic mean can be calculated if we do not have single attribute values but only grouped data. The reasons for grouping data have already been discussed: for instance, our banks group loans based on the principal amount into classes ranging from “micro-micro” to “large” because this is the only way to handle the thousands or millions of loans statistically.

When calculating the variance we face the same problem that already caused a bias of the arithmetic mean. In order to use the above formulas


with the frequencies, or the relative frequencies, respectively, we need one single value representing the whole class. The best choice would be the mean value of all class entries. The middle point of the class is usually not the mean value! However, in the absence of a better alternative we may also use the middle point km of class k.

We can now apply the above formulas for frequency tables by simply replacing the single attribute values kx by the reference value km :

• Grouped data with absolute frequencies:

2

21

1

1

( )1

( )

m

k k mk

k kmk

kk

f m x

VAR f m xn

f

=

=

=

⋅ −= = ⋅ −∑

∑∑

• Grouped data with relative frequencies:

2 2

1 1

( ) ( )m m

kk k k

k k

fVAR m x h m x

n= == ⋅ − = ⋅ −∑ ∑

EXAMPLE

We use the data of Tab. 7.7 with the calculated mean of 28.3333x = compared to the correct average of the complete list calculated with the Excel function AVERAGE = 29.167.


Pt. Abs. Rel. Calculations using Class M Freq. Freq. Squares Abs. Freq. Rel. Freq.

k km kf kh 2( )km x− 2( )k kf m x⋅ − 2( )k kh m x⋅ −

1 7.5 5 0.0833 434.03 2170.14 36.17

2 20 16 0.2667 69.44 1111.11 18.52

3 27.5 23 0.3833 0.69 15.97 0.27

4 35 7 0.1167 44.44 311.11 5.19

5 45 5 0.0833 277.78 1388.89 23.15

6 60 4 0.0667 1002.78 4011.11 66.85

Sum 60 1 9008.33 150.14

Table 7.12: Variance of grouped data

The calculations in the above are executed using the middle points. With the sum in the next-to-last column we find the variance:

9008.33150.14

60VAR= =

which is exactly the same value as was obtained by using the formula with the relative frequencies.

However, both values are based on the “middle point assumption”. We used the biased assumption for calculating the mean value, which is used afterwards for the variance. Therefore the variance is also biased.

The correct variance using the Excel function VARP on the complete list of the 60 numbers would be: 146.47VAR=

The alternative calculation of the variance (“mean of the squares minus square of the mean”) may be applied in an equivalent way:

2 2 2 2

1 1

1 m m

k k k kk k

VAR f m x h m xn = =

= ⋅ − = ⋅ −∑ ∑


The calculations will give exactly the same (biased) results as the original approach.

ATTENTION

In all calculations the biased mean value was smaller than the correct average: 28.333 as opposed to 29.167. This does not necessarily imply that the biased variance is automatically smaller. In fact, it is generally greater.

7.3.2 Standard Deviation

The variance is a figure which is directly derived from the attribute values of the statistics, i.e. it has basically the same dimension. However, it is the mean of a squared figure; therefore its dimension is also squared.

In order to illustrate this, let us look for Examples The statistic about loans consist of numbers with the dimension “currency” such as USD, EUR, or any other currency. The mean has the same dimension, i.e. the mean of all micro loans could be EUR 12433.45.

However, the variance would have the dimension “EUR-square”, because it is the mean of squared figures. Thus, we have to calculate its square root in order to obtain a quantity with the same dimension as the data which were at the origin of the calculations. The result is called standard deviation; it is the positive square root of the variance.

STANDARD DEVIATION

The positive square root of the variance is called standard deviation:

2

1

1( )

n

jj

xSD VA xRn =

−= = ∑

In statistics textbooks or in the lectures on statistics you may have attended in the past, you may have seen the small letter “s”, sometimes


also the small Greek letter sigma “σ ” used as symbols for the standard deviation.

Of course, if the variance is calculated using another method or formula, the above definition holds for the different formula as well. Independently of how the variance is calculated, its square root is always the standard deviation:

SD VAR=

EXAMPLES

1. The standard deviation of list 2L

2L : jx 2 7 4 -2 4 1 3 3 -2 4 5 6 -1 3 8

with the mean of 3x = and variance 12815

8.5333VAR= = is:

8.5333 2.9212SD VAR= = =

2. The calculations for grouped data are totally equivalent. The variance derived in Tab. 7.12 (see: Page 225) was:

9008.33150.14

60VAR= = → 150.14 12.25SD VAR= = =

EXCEL : In Excel we will again find the function for the standard deviation in the section “Statistical Functions”. In total you will find four different types, two of which are of interest to us: STDEV estimates the standard deviation for a sample. It is the square root of the VAR, i.e. the denominator of the mean calculation is n -1.

Since we are mainly operating with entire sets of data (= populations) we will usually deal with VARP, and therefore also with SDDEVP, which calculates the standard deviation for the entire population.


With the standard deviation we now have a performance figure with the same dimension as the attributes itself. Thus, if the attributes are the number of students, the standard deviation can be used as an indicator of how widely or narrowly the number of students scatters around the mean value. Thus, mean and standard deviation both have the dimension of the attribute values. Therefore, they can be added or subtracted.

In Fig. 7-14 we have marked two intervals:

• The first is called the “one-SD-interval” : AVE SD±

• The second is the “two- SD-interval” : 2AVE SD±

AVE1 SD1 SD2 SD 2 SD

AttributeValues

Figure 7-14: Spread around mean value

The SD tells us how far away numbers on a list are from their average. Most entries in the list will be somewhere around SD away from the average. We can expect to find many of the entries in the interval AVE SD± . Certainly many more – if not almost all – will lie in the interval 2AVE SD± which is twice as large as the first one.

This kind of interpretation of standard deviation in connection with the mean value is the most essential conclusion of the calculation.


EXAMPLE

We once more use the list of 60 entries given in Tab. 7.4 (see: Page 163):

25 15 16 30 26 5 20 30 24 37 21 25

25 17 19 25 28 26 52 26 30 26 21 45

49 22 21 43 26 27 19 29 32 27 30 12

66 13 23 29 27 39 29 28 35 27 40 48

57 1 24 28 28 33 27 29 38 26 63 41

At different steps of analysing the above statistical list we obtained:

• AVE = 29.1667

• VAR = 146.472

• SD = 12.1026

Following the above argumentations, we are interested to know how many numbers of the list will be in the one-SD-interval, and in the two-SD-interval, respectively.

• The one-SD-interval is [ ]17; 41 , i.e. we count the numbers of the

list in the range 17 41i≤ ≤ . The result is that 46 of the 60 numbers or 76.7% are in this range.

• The two-SD-interval is [5; 53], i.e. we look for numbers in the interval . We count 56 numbers or 93.3% in that range.

5 53i≤ ≤


VARIATION COEFFICIENT

Variance and standard deviation are measurement figures for spread; however, they cannot be interpreted in their absolute magnitude. The standard deviation may, for instance, be equal for micro loans and large loans. If the mean of the micro loans were EUR and the mean for the large loans were EUR, the spread of the large loans would be relatively much smaller because the values are spread around a large mean, whereas the spread around the micro average is relatively large. In order to avoid such a bias, statisticians have defined a coefficient which is the standard deviation relative to the mean. It is called the variation coefficient. Since SD and mean have the same dimension, the variation coefficient VC is dimension-free.

VARIATION COEFFICIENT

The variation coefficient is the quotient of standard deviation and the absolute value of the mean:

Assuming the same SD the variation coefficient would be 100 times larger in the case of micro loans compared to the large loans.

7.3.3 Normal Curve

If one measures the length of all newborn babies in a country, one will discover that the distribution of the lengths around the mean length of, say, 52 cm follows a curve which looks pretty much like the curve in Fig. 7-15.

2500m=250000l =

SDVC

x=


Rel

ativ

e Fr

eque

ncie

s

Distribution of Lengths

Size of Newborn Babies in cm

0.0%

5.0%

10.0%

15.0%

20.0%

25.0%

30.0%

45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Figure 7-15: Distribution of the sizes of newborn babies

It is amazing how many characteristics measured in nature follow this distribution, as for instance:

• the sum of n random numbers

• the weight of all 25-year-old men in a town

• the test score of the computer test of 100 applicants


Freqency Distribution

Sizes (grouped)

0

2%

4%

6%

8%

10%

12%

14%

16%

18%

20%

22%

161-165 166-170 171-173 186-190 191-195174-176 177-179 180-182 183-185

Perc

enta

ge

Figure 7-16: Histogram and fitted normal curve

If we draw a histogram of the sizes grouped in classes as shown on the horizontal scale of Fig. 7-16 we notice that the shape of the frequencies curve closely resembles the bell-shaped normal curve discovered by Abraham de Moivre from France (around 1720). Carl Friedrich Gauss (1777-1855), the German mathematician found the function describing the normal distribution.

It is an exponential function of the two parameters µ = average and σ =

standard deviation. The general function is:

The normal function involves three of the most famous numbers in the history of mathematics:

2 , π, and e

2121

2

x

y f ( x ) e

µσ

σ π

− − = =


The old 10 Deutschmark bill showed on the front side the portrait of C. F. Gauss together with the normal curve and the formula of the function.

Figure 7-17: The former DM 10 bill with normal murve

It is a remarkable fact that many histograms follow the normal curve. For such histograms, the average and the SD are good summary statistics.

If a histogram follows the normal curve, it looks something like the sketch in Fig. 7-17.


? ?????

-3 -2 -1 0 1 2 3

0

10

20

30

40

50

Pe

rce

nt

pe

r S

tan

da

rd U

nit

Standard Units

Problem Dependent Dimension

Normal Distribution with Standard Units

Figure 7-18: A normal-shaped histogram

STANDARD NORMAL DISTRIBUTION

Normalised to average and standard deviation (the Greek letters are used internationally for the average and SD!) the normal curve (bold line) has the following important properties:

• The graph is symmetric around 0, i.e. mean and median are 0.

• The total area under the curve is 1 (=100%) – as for every distribution curve. (Note the foregoing discussion about the area of the histogram!)

With respect to the horizontal scale of standard units (1 unit corresponds to 1 SD) we find:

• The area under the normal curve between −1 and +1 (= one-SD-interval) is about 68%.

• The area under the normal curve between −2 and +2 (= two-SD-interval) is about 95%.

• The area under the normal curve between −3 and +3 (= three-SD-interval,) is about 99.7%.

0µ = 1σ =


If the histogram for data is similar in shape to the normal curve, we could convert the horizontal scale to standard units by seeing how many SDs it is above (+ sign) or below (− sign) average.

For instance, take the data on student size, with an average of 178 cm and an SD of 6.02 cm. If we assume that the histogram is approximately bell-shaped, 178 cm is the most likely size in the class and approximately 68% of the students measure between , i.e. 172 ≤

≤ 184 in height. In fact, none of the students is exactly 178 cm tall,

but 42 = 70% of the students lie in the one-SD-range.

jx SD x x SD− ≤ ≤ +

jx


EXERCISE 7.3: KEY FIGURES: SPREAD

1. a) Calculate the variance of the numbers in the list with the mean :

1 -2 4 5 6 -3 0 3 7 -4 3 5 -6 -3 4 5 -3 4 8 5

b) Calculate the variance applying the alternative formula.

2. Given is the list of grades according to the German system (from 1 to 5) in the statistics exam. The mean grade was calculated to be

:

Student i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Grade 1 3 4 3 2 1 2 4 2 5 3 4 3 2 1 5 2 3 4 3

a) Calculate the variance of this list.

b) Use the frequency table compiled in exercise 7.2, problem 2 (see: Page 205) to verify the calculation of a) using the two alternative formulas (based on the absolute and the relative frequencies, respectively).

3. The following list of 70 grades was already used in exercise 7.1, problem 2 (see: Page 175).

2 2 4 3 2 1 1 4 2 5 3 4 3 2

1 1 2 4 3 3 3 5 4 2 5 3 1 3

5 1 1 4 3 2 4 3 5 1 2 2 3 4

3 3 4 2 5 1 2 1 1 2 4 4 3 5

3 2 3 2 4 5 2 1 4 3 5 2 3 2

1.95x =

2.85g =

jg


a) Compile a frequency table – with absolute and relative frequencies – using a tally list.

b) Calculate the variance of the above list using the simplified formulas.

c) Use Excel to determine the exact value of the variance and verify the calculations in b).

4. The following list contains 36 randomly generated numbers between 2 and 9:

3 8 2 3 4 8 9 3 8 7 5 5 9 7 5 7 8 8 2 6 3 6 2 7

6 6 7 5 3 8 2 2 5 7 6 7

a) Group the data into 3 classes: 2-4 (middle 3), 5-6 (middle 5.5), and 7-9 (middle 8). Compile a frequency table for these classes.

b) Calculate the variance for the grouped data using the simplified formula.

c) Determine the exact variance using Excel and explain the difference in the values.

5. Use the “Base-Problem” and calculate the variance using the Excel function VARP.

6. In exercise 7.1, problem 5 we created two frequency tables for the “Base-Problem” with different class widths. Calculate for both tables the variance using the simplified formula. Use Excel and organise the calculations in appropriate tables.

Compare the values with the exact value of problem 5.


7. Calculate the standard deviation for the data of the “Base-Problem”. You can either apply the corresponding Excel function, or use the variance from problem 6.

8. In exercise 7.1, problem 6 we created a histogram of the numbers of the “Base-Problem”. The shape is very close to the normal curve.

a) Determine the one-SD-interval around the mean value.

b) Decide from the sum curve the percentage of numbers inside the one-SD-range.

c) Use Excel to count the exact amount of numbers in the one-SD-range.

d) Compare these results with the theoretical figures derived from the normal distributions.


ANSWERS 7.3: KEY FIGURES: SPREAD

1. a) VAR = 16.1475

b) Same as a)

2. a) VAR = 1.4275

b) Same as a)

3. a)

Grade k kf kh 2

k kf g⋅ 2k kh g⋅

1 12 17.14% 12 0.171429

2 18 25.71% 72 1.028571

3 18 25.71% 162 2.314286

4 13 18.57% 208 2.971429

5 9 12.86% 225 3.214286

Sum 70

679 9.7

Absolute frequencies: 1.618163VAR=

Relative frequencies: same

b) same

4. a)

k Group kr km kf kh k kh m⋅ 2k kh m⋅

1 2-5 5 3 16 44.4% 1.33 4.00

2 5-6 6 5.5 5 13.9% 0.76 4.20

3 7-9 9 8 15 41.7% 3.33 26.67

Sum

36

5.43 34.87


b) 5.3771VAR=

c) 4.86034VARP=

5. VARP = 83.951

6. Class width 10: 95.33VAR=

Class width 5: 84.70VAR=

7. 9.1625SD=

8. a) 35.8735 54.1985jx≤ ≤

b) Approximately 64% = 320 numbers

c) With Excel: 305 numbers

d) Theoretical: 340 numbers


7.3.4 Progress Test “Key Figures: Spread”


At the end of the chapter you will find the solutions to the problems. Each solution is reduced to the final answer; it may even be only a number, a symbol, a table or a graph. You should check your answers. If they are correct, you have successfully completed this 2nd volume of the reader. Otherwise (if you got the wrong answer or no answer at all) you should repeat the corresponding units in order to close the gap.

1. Calculate the variance and the standard deviation for the following list of numbers with average = 2:

{0,7,3, 1, 1,4, 2,0,4,2.5, 4,3.5,1,4,1}L = − − −

2. For each list below, work out the average, the variance and the SD.

a) 1, 3, 4, 5, 7 b) 6, 8, 9, 10, 12

3. The average of the 40 numbers in problem 4 of exercise 7.1 is 26.05. After grouping the data we obtained the following frequency list.


Sum of rel. Fr.

17-19 19 18 3 7.5% 7.5%

20-22 22 21 8 20.0% 27.5%

23-25 25 24 10 25.0% 52.5%

26-28 28 27 7 17.5% 70.0%

29-31 31 30 5 12.5% 82.5%

32-34 34 33 4 10.0% 92.5%

35-37 37 36 3 7.5% 100.0%


Work out the variance and the standard deviation from the grouped data. Use the alternative (and simpler) formulas.

4. The management of your bank proposes to give all bank employees a flat raise of EUR 20 a month. What would this do to the average salary of the bank employees? What would it do to the SD?

5. Instead of raising all salaries by the same amount the management of your bank decides to grant a 5% increase in salaries across the board. What would this do to the average and the SD?

6. The following list of 25 test scores has an average of 50 and an SD of 10:

39 41 47 58 65 37 37 49 56 59

62 36 48 52 64 29 44 47 49 52

53 54 72 50 50

a) Use the normal approximation to estimate the number of scores within a one-SD-range of the average.

b) How many scores really are within the one-SD-range of the average?


7.4 Answers to Progress Tests

7.4.1 Answers to PT “Presentation Techniques”

You should check your answers. If they are correct, you may continue and start the next chapter. If not (if you got the wrong answer or no answer at all), repeat the topics of the recommended sections in order to close the gap.

1. 2 2

1 1

2 4n n

j jj j

x a x n a= =

− + ⋅∑ ∑

2. 41% for men; 51% for women.

3. a) Tally list:

No. Freq. Rel. F. No. Freq. Rel. F.

18 1 5% 24 3 15%

19 1 5% 25 1 5%

20 0 0% 26 2 10%

21 3 15% 27 1 5%

22 4 20% 28 1 5%

23 2 10% 29 1 5%

20 1


b) Column diagram:

0%

5%

10%

15%

20%

25%

18 19 20 21 22 23 24 25 26 27 28 29


Loans per LO

Rel

ativ

e F

requ

ency

4. Tally list:


Sum of rel. Fr.

17-19 19 18 3 7.5% 7.5%

20-22 22 21 8 20.0% 27.5%

23-25 25 24 10 25.0% 52.5%

26-28 28 27 7 17.5% 70.0%

29-31 31 30 5 12.5% 82.5%

32-34 34 33 4 10.0% 92.5%

35-37 37 36 3 7.5% 100.0%

40 100%


Column diagram of grouped data:


Group of Numbers

Rel

ativ

e F

requ

enci

es

0.0%

5.0%

10.0%

15.0%

20.0%

25.0%

30.0%

17-19 20-22 23-25 26-28 29-31 32-34 35-37

Sum curve → next problem 5.

5. Using the sum curve from problem 4, estimate the numbers between 24 and 30. How many numbers really were within this range?

0.0%

20.0%

40.0%

60.0%

80.0%

100.0%

120.0%

17-19 20-22 23-25 26-28 29-31 32-34 35-37

Sum Curve

Sum

of R

elat

ive

Fre

que

ncie

s

Group of Numbers

Estimate: 77% - 40% = 37% → 14 numbers

In fact, 12 numbers are really in this range.


7.4.2 Answers to PT “Key Figures: Centre”

You should check your answers. If they are correct, you may continue and start the next chapter. If not (if you got the wrong answer or no answer at all) repeat the topics of the recommended sections in order to close the gap.

1. Mean = 2; Median = 2.5; Mode = 4

2. 176.36 cm 3. 219 cm 4. 26.03

5. Graph with marked median and mean:

Group of Numbers

Sum

of R

elat

ive

Fre

quen

cies

Sum Curve

0%

20%

40%

60%

80%

100%

120%

17-19 20-22 23-25 26-28 29-31 32-34 35-37

Median

25

Mean

26

The distribution is right-tail.

A n s w e r s t o P r o g r e s s T e s t s P a g e | 247

7.4.3 Answers to PT “Key Figures: Spread”

You should check your answers. If they are correct, you have successfully completed this 2nd volume of the reader. If not (if you got the wrong answer or no answer at all) repeat the topics of the recommended sections in order to close the gap.

1. VAR = 5.9; SD = 2.43

2. a) AVE=4, VAR=4, SD=2 b) AVE=9, VAR=4, SD=2

3. VAR = 24.07; SD = 4.9

4. This would increase the average by EUR 20 but leave the SD unchanged.

5. This would increase the average and the SD by 5%.

6. a) One-SD 68% → 17 b) Really: 16

I n d e x P a g e | 249

Index

A

annuities .......................................... 87, 105

arithmetic mean ............................ 187, 188

attribute values ..................................... 188

average .................................................. 187

AVERAGE

function ............................................ 189

average value ........................................ 187

B

balance .................................................... 90

outstanding ....................................... 106

banking year ............................................ 59

base

logarithmic function ........................... 41

basic financial terms ................................ 62

basic rules ................................................ 85

bias ........................................................ 146

C

cardinal .................................................. 157

classification

data ................................................... 160

column diagram ..................................... 164

common logarithm ............................ 40, 43

compound factor ..................................... 61

compound interest ............................ 26, 60

compounding

continuously ....................................... 30

period ................................................. 68

periodically ......................................... 72

count index ............................................ 147

COUNTIF

function ............................................ 159

D

data

grouped ............................................. 161

decay

radioactive .......................................... 25

density distribution ................................ 166

deposits.................................................... 82

successive............................................ 84

diagram .................................................. 163

disbursement ......................................... 100

discount factor ......................................... 64

distribution

left-tail ............................................... 199

right-tail ............................................ 199

symmetric ......................................... 199

distribution curve ................................... 169

distribution function .............................. 170

domain

exponential function ........................... 20

E

effective interest rate ...................... 70, 113

exponential function ................................ 20

F

fees .......................................................... 70

frequency ............................................... 158

absolute ............................................ 161

relative ...................... 159, 161, 163, 165

table .......................................... 158, 163

FREQUENCY

function ............................................. 162

frequency sum ....................................... 167

frequency tables

grouped ............................................. 160

future value.............................................. 62

250 | P a g e I n d e x

G

geometric series ...................... 87, 107, 148

graph


logarithmic function ........................... 39

grouped data ......................................... 159

H

half-life .................................................... 25

histogram ...................................... 166, 196

normal .............................................. 231

I

in-payment .............................................. 54

instalment.............................. 100, 102, 105

interest .............................................. 26, 57

compound ........................................... 61

effective rate ...................................... 60

floating ................................................ 71

simple ........................................... 58, 83

interest days ............................................ 83

interest factor .................................... 27, 61

interest period ................................... 58, 68

interest rate ................................. 27, 58, 65

effective ................................ 69, 72, 113

nominal ................................. 68, 70, 113

periodic ............................................... 74

inverse function ....................................... 38

L

loan ........................................................ 100

logarithm ................................................. 40

logarithmic

operations........................................... 42

logarithmic function

properties ........................................... 40

M

maturity ................................................. 107

mean

normal curve ..................................... 232

mean value ............................................ 187

measure

spread ............................................... 214

median ........................................... 195, 196

normal curve ..................................... 232

middle point ........................... 161, 165, 193

mode ...................................................... 201

N

natural logarithm ..................................... 43

nominal .................................................. 157

normal curve .......................................... 230

normal distribution ................................ 229

number π ................................................ 28

number e .................................................. 28

number of periods ................................... 58

O

one dimensional list ............................... 157

one-SD-interval ...................................... 226

normal curve ..................................... 232

one-to-one function ................................. 39

ordinal .................................................... 157

out-payment ............................................ 54

outstanding amount ...................... 102, 111

P

payments

constant .............................................. 89

polynomial ............................................. 147

population ................ 14, 145, 161, 216, 225

population growth ............................. 15, 24

present value ..................................... 62, 63

principal ................................................... 58

principal repayment ....................... 102, 103

probability .............................................. 146

properties


logarithmic function ............................ 39

I n d e x P a g e | 251

R

radioactive

decay .................................................. 25

random variable .................................... 146

reference point .................................. 55, 85

reference value ...................................... 193

repayment ............................................. 101

repayment plan ..................................... 102

representation

graphical ........................................... 163

S

sample ................................... 145, 216, 225

savings ..................................................... 82

savings rate .............................................. 90

service charge .......................................... 70

spread .................................................... 211

standard deviation................................. 224

statistical features ................................. 189

statistical list .......................................... 188

statistical mean ..................................... 187

statistical sequence ............................... 157

sum curve ...................... 159, 163, 167, 169

sum symbol ........................................... 146

SUMPRODUCT

function ............................................ 192

T

tally list .................................................. 158

three-SD-interval

normal curve ..................................... 232

time scale ................................................ 54

time-line .................................................. 55

today-point ........................................ 55, 85

two-SD-interval ...................................... 226

normal curve ..................................... 232

V

variance ................................................. 215

alternative formula ........................... 218

frequencies ....................................... 219

grouped data ..................................... 222

unbiased ............................................ 216

variation coefficient ............................... 228

252 | P a g e N o t e s

N o t e s P a g e | 253

254 | P a g e N o t e s

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