unit 4 statistical methods btec nationals level 3

16
Mathematics for Engineering Technicians  299      U      N      I      T      4  Statistical Methods Your view of statistics has probably been formed from what you read in the papers, or what you see on the television. Survey use to show which political party is going to win the election, why men grow moustaches, if smoking damages your health, the average cost of housing by area, and all sorts of other interesting data! So statistics is used to analyse the results of such surveys and when used correctly, it attempts to eliminate the bias that often appears when collecting data on controve rsial issues. Statistics is concerned with collecting, sorting and analysing numerical facts, which originate from several observations. These facts are collated and summarized, then presented as tables, charts or diagrams, etc. In this brief introduction to statistics, we look at two specic areas. First, we consider the collection and presentation of data in its various forms. Then we look at how we measure such data, concentrating on nding average values. If you study statistics beyond this course, you will be introduced to the methods used to make predictions based on numerical data and the probability that your predictions are correct. At this stage in your learning, however, we will only be considering the areas of data handling and measurement of central tendency (averages), mentioned above. TYK 4.10 1.  A parallelogram h as an area of 60 cm 2  , if its perpendicular height is 10 cm, what is the length of one of the parallel sides? 2. Figure 4.43 shows the cross-section of a template, what is its area? 3.  An annulus has an inside diame ter of 0.75 m and an ext ernal diameter of 0.9 m, determine its area. 4. Find the volume of a circular cone of height 6 cm and base radius 5 cm. 5. Find the area of the curved surface of a cone (not including base) whose base radius is 3 cm and whose vertical height is 4 cm. Hint  : you need rst to nd the slant height. 6. If the area of a circle is 78.54 mm 2  , nd its diameter to 2 signicant gures. 7.  A cylinder of ba se radius 5 cm has a vo lume of 1 L (1000 cm 3   ), nd its height. 8.  A pipe of thick ness 5 mm has an e xternal diameter of 120 mm, nd the volume of 2.4 m of pipe material. 9.  A batch of 2000 ball bearings ar e each to hav e a diamete r of 5 mm. Determine the volume of metal needed for the manufacture of the whole batch. 10. Determine the volume and total surface area of a spherical shell having an internal diameter of 6 cm and external diameter of 8 cm. 50 cm 30 cm    6    0   c   m 3 cm 4 cm  Figure 4.43 Figure for question 2 in TYK 4.10        T     e    s    t   y o u  r      k     n       o    w  l    e   d    g  e TYK

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BTEC NATIONALS LEVEL 3, MATHS UNIT 4, STATISTICAL METHODS

TRANSCRIPT

Page 1: UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 29

Statistical Methods

Your view of statistics has probably been formed from what you read in

the papers or what you see on the television Survey use to show which

political party is going to win the election why men grow moustaches if

smoking damages your health the average cost of housing by area and all

sorts of other interesting data So statistics is used to analyse the results of

such surveys and when used correctly it attempts to eliminate the bias that

often appears when collecting data on controversial issues

Statistics is concerned with collecting sorting and analysing numerical

facts which originate from several observations These facts are collated

and summarized then presented as tables charts or diagrams etc

In this brief introduction to statistics we look at two specific areas First we

consider the collection and presentation of data in its various forms Then

we look at how we measure such data concentrating on finding average

values

If you study statistics beyond this course you will be introduced to

the methods used to make predictions based on numerical data and the

probability that your predictions are correct At this stage in your learning

however we will only be considering the areas of data handling and

measurement of central tendency (averages) mentioned above

TYK 410

1 A parallelogram has an area of 60 cm2 if its perpendicular height is 10 cm

what is the length of one of the parallel sides

2Figure 443 shows the cross-section of a template what is its area

3 An annulus has an inside diameter of 075 m and an external diameter of

09 m determine its area

4 Find the volume of a circular cone of height 6 cm and base radius 5 cm

5 Find the area of the curved surface of a cone (not including base) whose

base radius is 3 cm and whose vertical height is 4 cm Hint you need first

to find the slant height

6 If the area of a circle is 7854 mm2 find its diameter to 2 significant figures

7 A cylinder of base radius 5 cm has a volume of 1 L (1000 cm3 ) find its

height

8 A pipe of thickness 5 mm has an external diameter of 120 mm find thevolume of 24 m of pipe material

9 A batch of 2000 ball bearings are each to have a diameter of 5 mm

Determine the volume of metal needed for the manufacture of the whole

batch

10 Determine the volume and total surface area of a spherical shell having an

internal diameter of 6 cm and external diameter of 8 cm

50 cm

30 cm

6 0 c m

3 cm

4 cm

Figure 443 Figure for question 2 in TYK

410

T e s

t yo u r k

n

o w l e

d g e

TYK

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Mathematics for Engineering Technicians300

U N I T 4

Data manipulation

In almost all scientific engineering and business journals newspapers

and Government reports statistical information is presented in the form

of charts tables and diagrams as mentioned above We now look at a

small selection of these presentation methods including the necessary

manipulation of the data to produce them

Charts

Suppose as the result of a survey we are presented with the following

statistical data (Table 44 )

N u m b e r e m p l o y

e d

1000

800

600

400

200

0

P r i v a t e b u s i n e s s

P u b l i c b u s i n e s s

A g r i c u l t u r e

E n g i n e e r i n g

T r a n s p o r t

M a n u f a c t u r e

L e i s u r e i n d u s t r y

E d u c a t i o n

H e a l t h

O t h e r s

Category of employment

Figure 444 Bar chart representing number employed by category

KEY POINT

Statistics is concerned with collecting

sorting and analysing numerical facts

Table 44 Results of a survey

Major category of employment Number employed

Private business 750

Public business 900

Agriculture 200

Engineering 300

Transport 425

Manufacture 325

Leisure Industry 700

Education 775

Health 500

Other 125

Now ignoring for the moment the accuracy of this data let us look at

typical ways of presenting this information in the form of charts in

particular the bar chart and the pie chart

Bar chart

In its simplest form the bar chart may be used to represent data by drawing

individual bars (Figure 444 ) using the figures from the raw data (the data in

the table)

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 30

Now the scale for the vertical axis the number employed is easily decided

by considering the highest and lowest values in the table 900 and 125

respectively Therefore we use a scale from 0 to 1000 employees Along

the horizontal axis we represent each category by a bar of even width We

could just as easily have chosen to represent the data using column widths

instead of column heights

Now the simple bar chart above tells us very little that we could not have

determined from the table So another type of bar chart that enables us to

make comparisons the proportionate bar chart may be used

In this type of chart we use one bar with the same width throughout its

height with horizontal sections marked-off in proportion to the whole In

our example each section would represent the number of people employed

in each category compared with the total number of people surveyed

In order to draw a proportionate bar chart for our employment survey we

first need to total the number of people who took part in the survey This

total comes to 5000 Now even with this type of chart we may representthe data either in proportion by height or in proportion by percentage If

we were to choose height then we need to set our vertical scale at some

convenient height say 10 cm Then we would need to carry out 10 simple

calculations to determine the height of each individual column

For example given that the height of the total 10 cm represents 5000

people then the height of the column for those employed in private

business 750

500010 1 5

cm This type of calculation is then repeated

for each category of employment The resulting bar chart is shown in

Figure 445

10 cm Others

Health

Education

Leisure industry

Manufacture

Transport

Engineering

Agriculture

Public business

Private business

Figure 445 Proportionate bar chart graduated by height

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Mathematics for Engineering Technicians302

U N I T 4

Example 449

Draw a proportionate bar chart for the employment survey shown in Table 44

using the percentage method

For this method all that is required is to 1047297nd the appropriate percentage of the total

(5000) for each category of employment Then choosing a suitable height of column to

represent 100 mark on the appropriate percentage for each of the 10 employment

categories To save space only the 1047297rst 1047297ve categories of employment have been

calculated

1 private business 750

5000100 15

2 public business 900

5000100 18

3

agriculture 200

5000100 4

4 engineering 300

5000100 6

5 transport 425

5000100 8 5

Similarly manufacture 65 leisure industry 14 education 155 health 10

and other categories 25

Figure 446 shows the completed bar chart

Other categories of bar chart include horizontal bar charts where forinstance Figure 444 is turned through 90deg in a clockwise direction One

last type may be used to depict data given in chronological (time) order

Thus for example the horizontal x -axis is used to represent hours days

years etc while the vertical axis shows the variation of the data with time

Example 450

Represent the following data on a chronological bar chart

Year Number employed in general

engineering (thousands)

2003 800

2004 785

2005 690

2006 670

2007 590

Since we have not been asked to represent the data on any speci1047297c bar chart we will use

the simplest involving only the raw data Then the only concern is the scale we should

use for the vertical axis

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Mathematics for Engineering Technicians 30

To present a true representation the scale should start from zero and extend to say800 (Figure 447 a ) If we wish to emphasize a trend that is the way the variable is

rising or falling with time we could use a very much exaggerated scale (Figure 447 b )

This immediately emphasizes the downward trend since 1995 Note that this data is

1047297ctitious (made-up) and used here merely for emphasis

Pie chart

In this type of chart the data is presented as a proportion of the total using

the angle or area of sectors The method used to draw a pie chart is best

illustrated by example

Others (25)

Health (10)

Education (155)

Leisure industry (14)

Manufacture (65)

Transport (85)

Engineering (6)

Agriculture (4)

Public business (18)

Private business (15)

Figure 446 Proportionate percentage bar chart

N u m b e r e m p l o y e

d i n

e n g i n e e r i n g ( t h o u s

a n d s )

1000

900

800

700

600

500

400

300

200

100

02003 2004 2005 2006 2007

Time (years)

(a)

Figure 447 Chronological bar chart (a) in

correct proportion and (b) with graduated

scale

850

800

750

700

650

600

550

500

02003 2004 2005 2006 2007

Time (years)

(b)

N u m b e r e m p l o

y e d i n

e n g i n e e r i n g ( t h o

u s a n d s )

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Mathematics for Engineering Technicians304

U N I T 4

Example 451

Represent the data given in Example 450 on a pie chart

Remembering that there are 360deg in a circle and that the total number employed in

general engineering (according to our 1047297gures) was 800 785 690 670 590 3535

(thousands) then we manipulate the data as follows

Year Number employed in general

engineering (thousands)

Sector angle (to nearest half

degree)

2003 800

800

3535360 81 5

2004 785

785

3535360 80

2005 690

690

3535360 70 5

2006 670

670

3535360 68

2007 590

590

3535360 60

Total 3535 360deg

The resulting pie chart is shown in Figure 448

Other methods of visual presentation include pictograms and ideographs

These are diagrams in pictorial form used to present information to thosewho have a limited interest in the subject matter or who do not wish

to deal with data presented in numerical form They have little or no

practical use when interpreting engineering or other scientific data and

apart from acknowledging their existence we will not be pursuing them

further

Frequency distributions

One of the most common and most important ways of organizing and

presenting raw data is through use of frequency distributions

Consider the data given in Table 45 which shows the time in hours that it

took 50 individual workers to complete a specific assembly line task

2004

2005

2006

20072003

Figure 448 Resulting pie chart for

Example 451 employment in engineering

by year

Table 45 Data for assembly line task

11 10 06 11 09 11 08 09 12 07

10 15 09 14 10 09 11 10 10 11

08 09 12 07 06 12 09 08 07 10

10 12 10 10 11 14 07 11 09 09

08 11 10 10 13 05 08 13 13 08

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Mathematics for Engineering Technicians 30

From the data you should be able to see that the shortest time for completion

of the task was 05 hour the longest time was 15 hours The frequency of

appearance of these values is once On the other hand the number of times

the job took 1 hour appears 11 times or it has a frequency of 11 Trying

to sort out the data in this ad hoc manner is time consuming and may lead

to mistakes To assist with the task we use a tally chart This chart simply

shows how many times the event of completing the task in a specific time

takes place To record the frequency of events we use the number 1 in a

tally chart and when the frequency of the event reaches 5 we score through

the existing four 1rsquos to show a frequency of 5 The following example

illustrates the procedure

Example 452

Use a tally chart to determine the frequency of events for the data given on the

assembly line task in Table 45

Time (hours) Tally Frequency

05 1 1

06 11 2

07 1111 4

08 1111 1 6

09 1111 111 8

10 1111 1111 1 11

11 1111 111 8

12 1111 4

13 111 3

14 11 2

15 1 1

Total 50

We now have a full numerical representation of the frequency of events So for example

8 people completed the assembly task in 11 hours or the time 11 hours has a frequency

of 8 We will be using the above information later on when we consider measures of

central tendency

The times in hours given in the above data are simply numbers When data

appears in a form where it can be individually counted we say that it is

discrete data It goes up or down in countable steps Thus the numbers

12 34 86 9 111 130 are said to be discrete If however data is

obtained by measurement for example the heights of a group of people

then we say that this data is continuous When dealing with continuous

data we tend to quote its limits that is the limit of accuracy with which we

take the measurements So for example a person may be 174 05 cm

in height When dealing numerically with continuous data or a large

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Mathematics for Engineering Technicians306

U N I T 4

amount of discrete data it is often useful to group this data into classes or

categories We can then find out the numbers (frequency) of items within

each group

Table 46 shows the height of 200 adults grouped into 10 classes

KEY POINT

The grouping of frequency distributions

is a means for clearer presentation of

the facts

Table 46 Height of adults

Height (cm) Frequency

150ndash154 4

155ndash159 9

160ndash164 15

165ndash169 21

170ndash174 32

175ndash179 45

180ndash184 41

185ndash189 22

190ndash194 9

195ndash199 2

Total 200

The main advantage of grouping is that it produces a clear overall picture of

the frequency distribution In Table 46 the first class interval is 150ndash154

The end number 150 is known as the lower limit of the class interval and

the number 154 is the upper limit The heights have been measured to the

nearest centimetre That means within 05 cm Therefore in effect the

first class interval includes all heights in the range 1495ndash1545 cm these

numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and

upper class boundaries not the upper and lower limits of the class interval

The histogram and frequency graph

The histogram is a special diagram that is used to represent a frequency

distribution such as that for grouped heights shown above It consists of

a set of rectangles whose areas represent the frequencies of the various

classes Often when producing these diagrams the class width is kept the

same so that the varying frequencies are represented by the height of each

rectangle When drawing histograms for grouped data the midpoints of the

rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc

An adaptation of the histogram known as the frequency polygon may also

be used to represent a frequency distribution

Example 453

Represent the data shown in Table 46 on a histogram and draw in the frequency

polygon for this distribution

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Mathematics for Engineering Technicians 30

All that is required to produce the histogram is to plot frequency against the height

intervals where the intervals are drawn as class widths

Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects

the midpoint of the class widths

Another important method of representation is adding all the frequencies

of a distribution consecutively to produce a graph known as a cumulative

frequency distribution or Ogive

Figure 450 shows the cumulative frequency distribution graph for our data

given in Table 46 while Table 47 shows the consecutive addition of the

frequencies needed to produce the graph in Figure 450

From Figure 450 it is now a simple matter to find for example the median

grouped height or as it is more commonly known the 50th-percentile This

occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this

being in our case 100 giving an equivalent height of approximately 175 cm

Any percentile can be found for example the 75th-percentile where in

our case at a frequency of 150 the height can be seen to be approximately

180 cm

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

0

Class width 5 cm

Height of adults in cm

152 157 162 167 172 177 182 187 192 197

Figure 449 Figure for Example 453 histogram showing frequency distribution

KEY POINTThe frequencies of a distribution may

be added consecutively to produce a

graph known as a cumulative frequency

distribution

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Mathematics for Engineering Technicians308

U N I T 4

C u m u l a t i v e F r e q u e n c y

20

30

40

50

60

70

80

90

100

110

120

130140

150

160

170

180

190

200

10

0

152 157 162 167 172 177 182

180cm

187 192 197

75th percentile

50th percentile

Figure 450 Cumulative frequency distribution graph for data given in Table 46

Table 47 Cumulative frequency data

Height (cm) Frequency Cumulative frequency

150ndash154 4 4

155ndash159 9 13

160ndash164 15 28

165ndash169 21 49

170ndash174 32 81

175ndash179 45 126

180ndash184 41 167

185ndash189 22 189

190ndash194 9 198

195ndash199 2 200

Total 200 200

TYK 411

1 In a particular university the number of students enrolled by a faculty is

given in the table below

Faculty Number of students

Business and administration 1950

Humanities and social science 2820

Physical and life sciences 1050

Technology 850

Total 6670

T e s

t yo u r

k n o

w l e

d g e

TYK

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Mathematics for Engineering Technicians 30

Statistical measurement

When considering statistical data it is often convenient to have one or two

values that represent the data as a whole Average values are often used

You have already found an average value when looking at the median or

50th-percentile of a cumulative frequency distribution So for example we

might talk about the average height of females in the United Kingdom being

170 cm or that the average shoe size of British males is size 9 In statistics

we may represent these average values using the mean median or mode

of the data we are considering We will spend the rest of this short section

finding these average values for both discrete and grouped data starting

with the arithmetic mean

The arithmetic mean

The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the

numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up

and divide by how many there are or more formally

Arithmetic meanarithmetic total of all the individual val

uues

number of values

n

n

sum

where the Greek symbol the sum of the individual values

x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the

data

Illustrate this data on both a bar chart and pie chart

2 For the group of numbers given below produced a tally chart and

determine their frequency of occurrence

36 41 42 38 39 40 42 41 37 40

42 44 43 41 40 38 39 39 43 39

36 37 42 38 39 42 35 42 38 39

40 41 42 37 38 39 44 45 37 40

3 Given the following frequency distribution

Class interval Frequency ( f )

60ndash64 4

65ndash69 11

70ndash74 18

75ndash79 16

80ndash84 7

85ndash90 4

(a) produce a histogram and on it draw the frequency polygon

(b) produce a cumulative frequency graph and from it determine the value

of the 50th-percentile class

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Mathematics for Engineering Technicians310

U N I T 4

So for the mean of our ten numbers we have

mean

n

n

sum 8 7 9 10 5 6 12 9 6 8

10

80

108

Now no matter how long or complex the data we are dealing with provided

that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x

values rsquo is given the symbol x pronounced lsquo x bar rsquo

Example 454

The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm

163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find

the mean height of these females

Then for n 11

x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3

1118483

11168 03

x cm

Mean for grouped data

What if we are required to find the mean for grouped data Look back at

Table 46 showing the height of 200 adults grouped into ten classes In this

case the frequency of the heights needs to be taken into account

We select the class midpoint x as being the average of that class and then

multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the

frequency distribution the total value for the distribution is obtained ( fx )

This total is then divided by the sum of the frequencies ( f ) in order to

determine the mean So for grouped data

x f x f x f x f x

f f f f

f

f

n n

n

1 1 2 2 3 3

1 2 3

sdot

sdot

sumsum

( )midpoint

This rather complicated looking procedure is best illustrated by

example

Example 455

Determine the mean value for the heights of the 200 adults using the data in

Table 46

The values for each individual class are best found by producing a table using

the class midpoints and frequencies and remembering that the class midpoint is found by

dividing the sum of the upper and lower class boundaries by 2 So for example the mean

value for the 1047297rst class interval is149 5 154 5

2152

The completed table is shown

below

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Mathematics for Engineering Technicians 31

Midpoint ( x ) of height (cm) Frequency (f ) fx

152 4 608

157 9 1413

162 15 2430

167 21 3507

172 32 5504

177 45 7965

182 41 7462

187 22 4114

192 9 1728

197 2 394

Total f 200sum fx 35125sum

I hope you can see how each of the values was obtained Now that we have the required

totals the mean value of the distribution can be found

mean value x fx 35125

200175625 05 cm sum

sumf

Notice that our mean value of heights has the same margin of error as the original

measurements The value of the mean cannot be any more accurate than the measured

data from which it was found

Median

When some values within a set of data vary quite widely the arithmetic

mean gives a rather poor representative average of such data Under

these circumstances another more useful measure of the average is the

median

For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which

is not representative of any of the numbers given To find the median value

of the same set of numbers we simply place them in rank order that is 2

3 4 5 6 7 93 Then we select the middle (median) value Since there are

seven numbers (items) we choose the fourth item along the number 5 as

our median value

If the number of items in the set of values is even then we add together the

value of the two middle terms and divide by 2

Example 456

Find the mean and median value for the set of numbers 9 7 8 7 12 70 68

6 5 8

The arithmetic mean is found as

mean x

9 7 8 7 12 70 68 6 5 8

10

200

1020

This value is not really representative of any of the numbers in the set

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Mathematics for Engineering Technicians312

U N I T 4

To 1047297nd the median value we 1047297rst put the numbers in rank order that is

5 6 7 7 8 8 912 68 70

Then from the ten numbers the two middle values The 5th and 6th values along are 8

and 8 So the medianvalue

8 8

2 8

Mode

Yet another measure of central tendency for data containing extreme

values is the mode Now the mode of a set of values containing discrete

data is the value that occurs most often So for the set of values 4 4 4 5

5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs

four times Now it is possible for a set of data to have more than one

mode For example the data used in Example 462 above has two modes

7 and 8 both of these numbers occurring twice and both occurring more

than any of the others A set of data may not have a modal value at all For

example the numbers 2 3 4 5 6 7 8 all occur once and there is

no mode

A set of data that has one mode is called unimodal data with two

modes is bimodal and data with more than two modes is known as

multimodal

When considering frequency distributions for grouped data the modal

class is that group which occurs most frequently If we wish to find

the actual modal value of a frequency distribution we need to draw a

histogram

Example 457

Find the modal class and modal value for the frequency distribution of the height

of adults given in Table 46

Referring back to Table 46 it is easy to see that the class of heights which occurs most

frequently is 175 ndash 179 cm which occurs 45 times

Now to 1047297nd the modal value we need to produce a histogram for the data

We did this for Example 453 This histogram is shown again here with the modalshown

From Figure 451 it can be seen that the modal value 17825 05 cm

This value is obtained from the intersection of the two construction lines AB and CD The

line AB is drawn diagonally from the highest value of the preceding class up to the top

right-hand corner of the modal class The line CD is drawn from the top left-hand corner

of the modal group to the lowest value of the next class immediately above the modal

group Then as can be seen the modal value is read-off where the projection line meets

the x -axis

KEY POINT

The mean median and mode are

statistical averages or measures

of central tendency for a statistical

distribution

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Mathematics for Engineering Technicians 31

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

C

A

B

D

0Height of adults (cm)

152 157 162 167 172 177 182 187 192 197

Modal value 17825 05 cm

Figure 451 Histogram showing frequency distribution and modal value for height of adults

TYK 412

1 Calculate the mean of the numbers 1765 986 1124 1898 959 and

888

2 Determine the mean the median and the mode for the set of numbers 9 8

7 27 16 3 1 9 4 and 116

3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the

arithmetic mean the median and the mode

4 Estimates for the length of wood required for a shelf were as follows

Length (cm) 35 36 37 38 39 40 41 42

Frequency 1 3 4 8 6 5 3 2

Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table

Length (mm) 167 168 169 170 171

Frequency 2 7 20 8 3

6 Calculate the arithmetic mean for the data shown in the table

Length of rivet (mm) 98 99 995 100 1005 101 102

Frequency 3 18 36 62 56 20 5

T e s t yo u

r k

n o

w l e

d g

e

TYK

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Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present

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Mathematics for Engineering Technicians300

U N I T 4

Data manipulation

In almost all scientific engineering and business journals newspapers

and Government reports statistical information is presented in the form

of charts tables and diagrams as mentioned above We now look at a

small selection of these presentation methods including the necessary

manipulation of the data to produce them

Charts

Suppose as the result of a survey we are presented with the following

statistical data (Table 44 )

N u m b e r e m p l o y

e d

1000

800

600

400

200

0

P r i v a t e b u s i n e s s

P u b l i c b u s i n e s s

A g r i c u l t u r e

E n g i n e e r i n g

T r a n s p o r t

M a n u f a c t u r e

L e i s u r e i n d u s t r y

E d u c a t i o n

H e a l t h

O t h e r s

Category of employment

Figure 444 Bar chart representing number employed by category

KEY POINT

Statistics is concerned with collecting

sorting and analysing numerical facts

Table 44 Results of a survey

Major category of employment Number employed

Private business 750

Public business 900

Agriculture 200

Engineering 300

Transport 425

Manufacture 325

Leisure Industry 700

Education 775

Health 500

Other 125

Now ignoring for the moment the accuracy of this data let us look at

typical ways of presenting this information in the form of charts in

particular the bar chart and the pie chart

Bar chart

In its simplest form the bar chart may be used to represent data by drawing

individual bars (Figure 444 ) using the figures from the raw data (the data in

the table)

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Mathematics for Engineering Technicians 30

Now the scale for the vertical axis the number employed is easily decided

by considering the highest and lowest values in the table 900 and 125

respectively Therefore we use a scale from 0 to 1000 employees Along

the horizontal axis we represent each category by a bar of even width We

could just as easily have chosen to represent the data using column widths

instead of column heights

Now the simple bar chart above tells us very little that we could not have

determined from the table So another type of bar chart that enables us to

make comparisons the proportionate bar chart may be used

In this type of chart we use one bar with the same width throughout its

height with horizontal sections marked-off in proportion to the whole In

our example each section would represent the number of people employed

in each category compared with the total number of people surveyed

In order to draw a proportionate bar chart for our employment survey we

first need to total the number of people who took part in the survey This

total comes to 5000 Now even with this type of chart we may representthe data either in proportion by height or in proportion by percentage If

we were to choose height then we need to set our vertical scale at some

convenient height say 10 cm Then we would need to carry out 10 simple

calculations to determine the height of each individual column

For example given that the height of the total 10 cm represents 5000

people then the height of the column for those employed in private

business 750

500010 1 5

cm This type of calculation is then repeated

for each category of employment The resulting bar chart is shown in

Figure 445

10 cm Others

Health

Education

Leisure industry

Manufacture

Transport

Engineering

Agriculture

Public business

Private business

Figure 445 Proportionate bar chart graduated by height

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Mathematics for Engineering Technicians302

U N I T 4

Example 449

Draw a proportionate bar chart for the employment survey shown in Table 44

using the percentage method

For this method all that is required is to 1047297nd the appropriate percentage of the total

(5000) for each category of employment Then choosing a suitable height of column to

represent 100 mark on the appropriate percentage for each of the 10 employment

categories To save space only the 1047297rst 1047297ve categories of employment have been

calculated

1 private business 750

5000100 15

2 public business 900

5000100 18

3

agriculture 200

5000100 4

4 engineering 300

5000100 6

5 transport 425

5000100 8 5

Similarly manufacture 65 leisure industry 14 education 155 health 10

and other categories 25

Figure 446 shows the completed bar chart

Other categories of bar chart include horizontal bar charts where forinstance Figure 444 is turned through 90deg in a clockwise direction One

last type may be used to depict data given in chronological (time) order

Thus for example the horizontal x -axis is used to represent hours days

years etc while the vertical axis shows the variation of the data with time

Example 450

Represent the following data on a chronological bar chart

Year Number employed in general

engineering (thousands)

2003 800

2004 785

2005 690

2006 670

2007 590

Since we have not been asked to represent the data on any speci1047297c bar chart we will use

the simplest involving only the raw data Then the only concern is the scale we should

use for the vertical axis

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Mathematics for Engineering Technicians 30

To present a true representation the scale should start from zero and extend to say800 (Figure 447 a ) If we wish to emphasize a trend that is the way the variable is

rising or falling with time we could use a very much exaggerated scale (Figure 447 b )

This immediately emphasizes the downward trend since 1995 Note that this data is

1047297ctitious (made-up) and used here merely for emphasis

Pie chart

In this type of chart the data is presented as a proportion of the total using

the angle or area of sectors The method used to draw a pie chart is best

illustrated by example

Others (25)

Health (10)

Education (155)

Leisure industry (14)

Manufacture (65)

Transport (85)

Engineering (6)

Agriculture (4)

Public business (18)

Private business (15)

Figure 446 Proportionate percentage bar chart

N u m b e r e m p l o y e

d i n

e n g i n e e r i n g ( t h o u s

a n d s )

1000

900

800

700

600

500

400

300

200

100

02003 2004 2005 2006 2007

Time (years)

(a)

Figure 447 Chronological bar chart (a) in

correct proportion and (b) with graduated

scale

850

800

750

700

650

600

550

500

02003 2004 2005 2006 2007

Time (years)

(b)

N u m b e r e m p l o

y e d i n

e n g i n e e r i n g ( t h o

u s a n d s )

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Mathematics for Engineering Technicians304

U N I T 4

Example 451

Represent the data given in Example 450 on a pie chart

Remembering that there are 360deg in a circle and that the total number employed in

general engineering (according to our 1047297gures) was 800 785 690 670 590 3535

(thousands) then we manipulate the data as follows

Year Number employed in general

engineering (thousands)

Sector angle (to nearest half

degree)

2003 800

800

3535360 81 5

2004 785

785

3535360 80

2005 690

690

3535360 70 5

2006 670

670

3535360 68

2007 590

590

3535360 60

Total 3535 360deg

The resulting pie chart is shown in Figure 448

Other methods of visual presentation include pictograms and ideographs

These are diagrams in pictorial form used to present information to thosewho have a limited interest in the subject matter or who do not wish

to deal with data presented in numerical form They have little or no

practical use when interpreting engineering or other scientific data and

apart from acknowledging their existence we will not be pursuing them

further

Frequency distributions

One of the most common and most important ways of organizing and

presenting raw data is through use of frequency distributions

Consider the data given in Table 45 which shows the time in hours that it

took 50 individual workers to complete a specific assembly line task

2004

2005

2006

20072003

Figure 448 Resulting pie chart for

Example 451 employment in engineering

by year

Table 45 Data for assembly line task

11 10 06 11 09 11 08 09 12 07

10 15 09 14 10 09 11 10 10 11

08 09 12 07 06 12 09 08 07 10

10 12 10 10 11 14 07 11 09 09

08 11 10 10 13 05 08 13 13 08

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Mathematics for Engineering Technicians 30

From the data you should be able to see that the shortest time for completion

of the task was 05 hour the longest time was 15 hours The frequency of

appearance of these values is once On the other hand the number of times

the job took 1 hour appears 11 times or it has a frequency of 11 Trying

to sort out the data in this ad hoc manner is time consuming and may lead

to mistakes To assist with the task we use a tally chart This chart simply

shows how many times the event of completing the task in a specific time

takes place To record the frequency of events we use the number 1 in a

tally chart and when the frequency of the event reaches 5 we score through

the existing four 1rsquos to show a frequency of 5 The following example

illustrates the procedure

Example 452

Use a tally chart to determine the frequency of events for the data given on the

assembly line task in Table 45

Time (hours) Tally Frequency

05 1 1

06 11 2

07 1111 4

08 1111 1 6

09 1111 111 8

10 1111 1111 1 11

11 1111 111 8

12 1111 4

13 111 3

14 11 2

15 1 1

Total 50

We now have a full numerical representation of the frequency of events So for example

8 people completed the assembly task in 11 hours or the time 11 hours has a frequency

of 8 We will be using the above information later on when we consider measures of

central tendency

The times in hours given in the above data are simply numbers When data

appears in a form where it can be individually counted we say that it is

discrete data It goes up or down in countable steps Thus the numbers

12 34 86 9 111 130 are said to be discrete If however data is

obtained by measurement for example the heights of a group of people

then we say that this data is continuous When dealing with continuous

data we tend to quote its limits that is the limit of accuracy with which we

take the measurements So for example a person may be 174 05 cm

in height When dealing numerically with continuous data or a large

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Mathematics for Engineering Technicians306

U N I T 4

amount of discrete data it is often useful to group this data into classes or

categories We can then find out the numbers (frequency) of items within

each group

Table 46 shows the height of 200 adults grouped into 10 classes

KEY POINT

The grouping of frequency distributions

is a means for clearer presentation of

the facts

Table 46 Height of adults

Height (cm) Frequency

150ndash154 4

155ndash159 9

160ndash164 15

165ndash169 21

170ndash174 32

175ndash179 45

180ndash184 41

185ndash189 22

190ndash194 9

195ndash199 2

Total 200

The main advantage of grouping is that it produces a clear overall picture of

the frequency distribution In Table 46 the first class interval is 150ndash154

The end number 150 is known as the lower limit of the class interval and

the number 154 is the upper limit The heights have been measured to the

nearest centimetre That means within 05 cm Therefore in effect the

first class interval includes all heights in the range 1495ndash1545 cm these

numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and

upper class boundaries not the upper and lower limits of the class interval

The histogram and frequency graph

The histogram is a special diagram that is used to represent a frequency

distribution such as that for grouped heights shown above It consists of

a set of rectangles whose areas represent the frequencies of the various

classes Often when producing these diagrams the class width is kept the

same so that the varying frequencies are represented by the height of each

rectangle When drawing histograms for grouped data the midpoints of the

rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc

An adaptation of the histogram known as the frequency polygon may also

be used to represent a frequency distribution

Example 453

Represent the data shown in Table 46 on a histogram and draw in the frequency

polygon for this distribution

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Mathematics for Engineering Technicians 30

All that is required to produce the histogram is to plot frequency against the height

intervals where the intervals are drawn as class widths

Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects

the midpoint of the class widths

Another important method of representation is adding all the frequencies

of a distribution consecutively to produce a graph known as a cumulative

frequency distribution or Ogive

Figure 450 shows the cumulative frequency distribution graph for our data

given in Table 46 while Table 47 shows the consecutive addition of the

frequencies needed to produce the graph in Figure 450

From Figure 450 it is now a simple matter to find for example the median

grouped height or as it is more commonly known the 50th-percentile This

occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this

being in our case 100 giving an equivalent height of approximately 175 cm

Any percentile can be found for example the 75th-percentile where in

our case at a frequency of 150 the height can be seen to be approximately

180 cm

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

0

Class width 5 cm

Height of adults in cm

152 157 162 167 172 177 182 187 192 197

Figure 449 Figure for Example 453 histogram showing frequency distribution

KEY POINTThe frequencies of a distribution may

be added consecutively to produce a

graph known as a cumulative frequency

distribution

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Mathematics for Engineering Technicians308

U N I T 4

C u m u l a t i v e F r e q u e n c y

20

30

40

50

60

70

80

90

100

110

120

130140

150

160

170

180

190

200

10

0

152 157 162 167 172 177 182

180cm

187 192 197

75th percentile

50th percentile

Figure 450 Cumulative frequency distribution graph for data given in Table 46

Table 47 Cumulative frequency data

Height (cm) Frequency Cumulative frequency

150ndash154 4 4

155ndash159 9 13

160ndash164 15 28

165ndash169 21 49

170ndash174 32 81

175ndash179 45 126

180ndash184 41 167

185ndash189 22 189

190ndash194 9 198

195ndash199 2 200

Total 200 200

TYK 411

1 In a particular university the number of students enrolled by a faculty is

given in the table below

Faculty Number of students

Business and administration 1950

Humanities and social science 2820

Physical and life sciences 1050

Technology 850

Total 6670

T e s

t yo u r

k n o

w l e

d g e

TYK

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Mathematics for Engineering Technicians 30

Statistical measurement

When considering statistical data it is often convenient to have one or two

values that represent the data as a whole Average values are often used

You have already found an average value when looking at the median or

50th-percentile of a cumulative frequency distribution So for example we

might talk about the average height of females in the United Kingdom being

170 cm or that the average shoe size of British males is size 9 In statistics

we may represent these average values using the mean median or mode

of the data we are considering We will spend the rest of this short section

finding these average values for both discrete and grouped data starting

with the arithmetic mean

The arithmetic mean

The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the

numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up

and divide by how many there are or more formally

Arithmetic meanarithmetic total of all the individual val

uues

number of values

n

n

sum

where the Greek symbol the sum of the individual values

x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the

data

Illustrate this data on both a bar chart and pie chart

2 For the group of numbers given below produced a tally chart and

determine their frequency of occurrence

36 41 42 38 39 40 42 41 37 40

42 44 43 41 40 38 39 39 43 39

36 37 42 38 39 42 35 42 38 39

40 41 42 37 38 39 44 45 37 40

3 Given the following frequency distribution

Class interval Frequency ( f )

60ndash64 4

65ndash69 11

70ndash74 18

75ndash79 16

80ndash84 7

85ndash90 4

(a) produce a histogram and on it draw the frequency polygon

(b) produce a cumulative frequency graph and from it determine the value

of the 50th-percentile class

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Mathematics for Engineering Technicians310

U N I T 4

So for the mean of our ten numbers we have

mean

n

n

sum 8 7 9 10 5 6 12 9 6 8

10

80

108

Now no matter how long or complex the data we are dealing with provided

that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x

values rsquo is given the symbol x pronounced lsquo x bar rsquo

Example 454

The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm

163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find

the mean height of these females

Then for n 11

x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3

1118483

11168 03

x cm

Mean for grouped data

What if we are required to find the mean for grouped data Look back at

Table 46 showing the height of 200 adults grouped into ten classes In this

case the frequency of the heights needs to be taken into account

We select the class midpoint x as being the average of that class and then

multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the

frequency distribution the total value for the distribution is obtained ( fx )

This total is then divided by the sum of the frequencies ( f ) in order to

determine the mean So for grouped data

x f x f x f x f x

f f f f

f

f

n n

n

1 1 2 2 3 3

1 2 3

sdot

sdot

sumsum

( )midpoint

This rather complicated looking procedure is best illustrated by

example

Example 455

Determine the mean value for the heights of the 200 adults using the data in

Table 46

The values for each individual class are best found by producing a table using

the class midpoints and frequencies and remembering that the class midpoint is found by

dividing the sum of the upper and lower class boundaries by 2 So for example the mean

value for the 1047297rst class interval is149 5 154 5

2152

The completed table is shown

below

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Mathematics for Engineering Technicians 31

Midpoint ( x ) of height (cm) Frequency (f ) fx

152 4 608

157 9 1413

162 15 2430

167 21 3507

172 32 5504

177 45 7965

182 41 7462

187 22 4114

192 9 1728

197 2 394

Total f 200sum fx 35125sum

I hope you can see how each of the values was obtained Now that we have the required

totals the mean value of the distribution can be found

mean value x fx 35125

200175625 05 cm sum

sumf

Notice that our mean value of heights has the same margin of error as the original

measurements The value of the mean cannot be any more accurate than the measured

data from which it was found

Median

When some values within a set of data vary quite widely the arithmetic

mean gives a rather poor representative average of such data Under

these circumstances another more useful measure of the average is the

median

For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which

is not representative of any of the numbers given To find the median value

of the same set of numbers we simply place them in rank order that is 2

3 4 5 6 7 93 Then we select the middle (median) value Since there are

seven numbers (items) we choose the fourth item along the number 5 as

our median value

If the number of items in the set of values is even then we add together the

value of the two middle terms and divide by 2

Example 456

Find the mean and median value for the set of numbers 9 7 8 7 12 70 68

6 5 8

The arithmetic mean is found as

mean x

9 7 8 7 12 70 68 6 5 8

10

200

1020

This value is not really representative of any of the numbers in the set

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Mathematics for Engineering Technicians312

U N I T 4

To 1047297nd the median value we 1047297rst put the numbers in rank order that is

5 6 7 7 8 8 912 68 70

Then from the ten numbers the two middle values The 5th and 6th values along are 8

and 8 So the medianvalue

8 8

2 8

Mode

Yet another measure of central tendency for data containing extreme

values is the mode Now the mode of a set of values containing discrete

data is the value that occurs most often So for the set of values 4 4 4 5

5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs

four times Now it is possible for a set of data to have more than one

mode For example the data used in Example 462 above has two modes

7 and 8 both of these numbers occurring twice and both occurring more

than any of the others A set of data may not have a modal value at all For

example the numbers 2 3 4 5 6 7 8 all occur once and there is

no mode

A set of data that has one mode is called unimodal data with two

modes is bimodal and data with more than two modes is known as

multimodal

When considering frequency distributions for grouped data the modal

class is that group which occurs most frequently If we wish to find

the actual modal value of a frequency distribution we need to draw a

histogram

Example 457

Find the modal class and modal value for the frequency distribution of the height

of adults given in Table 46

Referring back to Table 46 it is easy to see that the class of heights which occurs most

frequently is 175 ndash 179 cm which occurs 45 times

Now to 1047297nd the modal value we need to produce a histogram for the data

We did this for Example 453 This histogram is shown again here with the modalshown

From Figure 451 it can be seen that the modal value 17825 05 cm

This value is obtained from the intersection of the two construction lines AB and CD The

line AB is drawn diagonally from the highest value of the preceding class up to the top

right-hand corner of the modal class The line CD is drawn from the top left-hand corner

of the modal group to the lowest value of the next class immediately above the modal

group Then as can be seen the modal value is read-off where the projection line meets

the x -axis

KEY POINT

The mean median and mode are

statistical averages or measures

of central tendency for a statistical

distribution

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Mathematics for Engineering Technicians 31

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

C

A

B

D

0Height of adults (cm)

152 157 162 167 172 177 182 187 192 197

Modal value 17825 05 cm

Figure 451 Histogram showing frequency distribution and modal value for height of adults

TYK 412

1 Calculate the mean of the numbers 1765 986 1124 1898 959 and

888

2 Determine the mean the median and the mode for the set of numbers 9 8

7 27 16 3 1 9 4 and 116

3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the

arithmetic mean the median and the mode

4 Estimates for the length of wood required for a shelf were as follows

Length (cm) 35 36 37 38 39 40 41 42

Frequency 1 3 4 8 6 5 3 2

Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table

Length (mm) 167 168 169 170 171

Frequency 2 7 20 8 3

6 Calculate the arithmetic mean for the data shown in the table

Length of rivet (mm) 98 99 995 100 1005 101 102

Frequency 3 18 36 62 56 20 5

T e s t yo u

r k

n o

w l e

d g

e

TYK

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Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present

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Mathematics for Engineering Technicians 30

Now the scale for the vertical axis the number employed is easily decided

by considering the highest and lowest values in the table 900 and 125

respectively Therefore we use a scale from 0 to 1000 employees Along

the horizontal axis we represent each category by a bar of even width We

could just as easily have chosen to represent the data using column widths

instead of column heights

Now the simple bar chart above tells us very little that we could not have

determined from the table So another type of bar chart that enables us to

make comparisons the proportionate bar chart may be used

In this type of chart we use one bar with the same width throughout its

height with horizontal sections marked-off in proportion to the whole In

our example each section would represent the number of people employed

in each category compared with the total number of people surveyed

In order to draw a proportionate bar chart for our employment survey we

first need to total the number of people who took part in the survey This

total comes to 5000 Now even with this type of chart we may representthe data either in proportion by height or in proportion by percentage If

we were to choose height then we need to set our vertical scale at some

convenient height say 10 cm Then we would need to carry out 10 simple

calculations to determine the height of each individual column

For example given that the height of the total 10 cm represents 5000

people then the height of the column for those employed in private

business 750

500010 1 5

cm This type of calculation is then repeated

for each category of employment The resulting bar chart is shown in

Figure 445

10 cm Others

Health

Education

Leisure industry

Manufacture

Transport

Engineering

Agriculture

Public business

Private business

Figure 445 Proportionate bar chart graduated by height

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Mathematics for Engineering Technicians302

U N I T 4

Example 449

Draw a proportionate bar chart for the employment survey shown in Table 44

using the percentage method

For this method all that is required is to 1047297nd the appropriate percentage of the total

(5000) for each category of employment Then choosing a suitable height of column to

represent 100 mark on the appropriate percentage for each of the 10 employment

categories To save space only the 1047297rst 1047297ve categories of employment have been

calculated

1 private business 750

5000100 15

2 public business 900

5000100 18

3

agriculture 200

5000100 4

4 engineering 300

5000100 6

5 transport 425

5000100 8 5

Similarly manufacture 65 leisure industry 14 education 155 health 10

and other categories 25

Figure 446 shows the completed bar chart

Other categories of bar chart include horizontal bar charts where forinstance Figure 444 is turned through 90deg in a clockwise direction One

last type may be used to depict data given in chronological (time) order

Thus for example the horizontal x -axis is used to represent hours days

years etc while the vertical axis shows the variation of the data with time

Example 450

Represent the following data on a chronological bar chart

Year Number employed in general

engineering (thousands)

2003 800

2004 785

2005 690

2006 670

2007 590

Since we have not been asked to represent the data on any speci1047297c bar chart we will use

the simplest involving only the raw data Then the only concern is the scale we should

use for the vertical axis

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Mathematics for Engineering Technicians 30

To present a true representation the scale should start from zero and extend to say800 (Figure 447 a ) If we wish to emphasize a trend that is the way the variable is

rising or falling with time we could use a very much exaggerated scale (Figure 447 b )

This immediately emphasizes the downward trend since 1995 Note that this data is

1047297ctitious (made-up) and used here merely for emphasis

Pie chart

In this type of chart the data is presented as a proportion of the total using

the angle or area of sectors The method used to draw a pie chart is best

illustrated by example

Others (25)

Health (10)

Education (155)

Leisure industry (14)

Manufacture (65)

Transport (85)

Engineering (6)

Agriculture (4)

Public business (18)

Private business (15)

Figure 446 Proportionate percentage bar chart

N u m b e r e m p l o y e

d i n

e n g i n e e r i n g ( t h o u s

a n d s )

1000

900

800

700

600

500

400

300

200

100

02003 2004 2005 2006 2007

Time (years)

(a)

Figure 447 Chronological bar chart (a) in

correct proportion and (b) with graduated

scale

850

800

750

700

650

600

550

500

02003 2004 2005 2006 2007

Time (years)

(b)

N u m b e r e m p l o

y e d i n

e n g i n e e r i n g ( t h o

u s a n d s )

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Mathematics for Engineering Technicians304

U N I T 4

Example 451

Represent the data given in Example 450 on a pie chart

Remembering that there are 360deg in a circle and that the total number employed in

general engineering (according to our 1047297gures) was 800 785 690 670 590 3535

(thousands) then we manipulate the data as follows

Year Number employed in general

engineering (thousands)

Sector angle (to nearest half

degree)

2003 800

800

3535360 81 5

2004 785

785

3535360 80

2005 690

690

3535360 70 5

2006 670

670

3535360 68

2007 590

590

3535360 60

Total 3535 360deg

The resulting pie chart is shown in Figure 448

Other methods of visual presentation include pictograms and ideographs

These are diagrams in pictorial form used to present information to thosewho have a limited interest in the subject matter or who do not wish

to deal with data presented in numerical form They have little or no

practical use when interpreting engineering or other scientific data and

apart from acknowledging their existence we will not be pursuing them

further

Frequency distributions

One of the most common and most important ways of organizing and

presenting raw data is through use of frequency distributions

Consider the data given in Table 45 which shows the time in hours that it

took 50 individual workers to complete a specific assembly line task

2004

2005

2006

20072003

Figure 448 Resulting pie chart for

Example 451 employment in engineering

by year

Table 45 Data for assembly line task

11 10 06 11 09 11 08 09 12 07

10 15 09 14 10 09 11 10 10 11

08 09 12 07 06 12 09 08 07 10

10 12 10 10 11 14 07 11 09 09

08 11 10 10 13 05 08 13 13 08

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Mathematics for Engineering Technicians 30

From the data you should be able to see that the shortest time for completion

of the task was 05 hour the longest time was 15 hours The frequency of

appearance of these values is once On the other hand the number of times

the job took 1 hour appears 11 times or it has a frequency of 11 Trying

to sort out the data in this ad hoc manner is time consuming and may lead

to mistakes To assist with the task we use a tally chart This chart simply

shows how many times the event of completing the task in a specific time

takes place To record the frequency of events we use the number 1 in a

tally chart and when the frequency of the event reaches 5 we score through

the existing four 1rsquos to show a frequency of 5 The following example

illustrates the procedure

Example 452

Use a tally chart to determine the frequency of events for the data given on the

assembly line task in Table 45

Time (hours) Tally Frequency

05 1 1

06 11 2

07 1111 4

08 1111 1 6

09 1111 111 8

10 1111 1111 1 11

11 1111 111 8

12 1111 4

13 111 3

14 11 2

15 1 1

Total 50

We now have a full numerical representation of the frequency of events So for example

8 people completed the assembly task in 11 hours or the time 11 hours has a frequency

of 8 We will be using the above information later on when we consider measures of

central tendency

The times in hours given in the above data are simply numbers When data

appears in a form where it can be individually counted we say that it is

discrete data It goes up or down in countable steps Thus the numbers

12 34 86 9 111 130 are said to be discrete If however data is

obtained by measurement for example the heights of a group of people

then we say that this data is continuous When dealing with continuous

data we tend to quote its limits that is the limit of accuracy with which we

take the measurements So for example a person may be 174 05 cm

in height When dealing numerically with continuous data or a large

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Mathematics for Engineering Technicians306

U N I T 4

amount of discrete data it is often useful to group this data into classes or

categories We can then find out the numbers (frequency) of items within

each group

Table 46 shows the height of 200 adults grouped into 10 classes

KEY POINT

The grouping of frequency distributions

is a means for clearer presentation of

the facts

Table 46 Height of adults

Height (cm) Frequency

150ndash154 4

155ndash159 9

160ndash164 15

165ndash169 21

170ndash174 32

175ndash179 45

180ndash184 41

185ndash189 22

190ndash194 9

195ndash199 2

Total 200

The main advantage of grouping is that it produces a clear overall picture of

the frequency distribution In Table 46 the first class interval is 150ndash154

The end number 150 is known as the lower limit of the class interval and

the number 154 is the upper limit The heights have been measured to the

nearest centimetre That means within 05 cm Therefore in effect the

first class interval includes all heights in the range 1495ndash1545 cm these

numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and

upper class boundaries not the upper and lower limits of the class interval

The histogram and frequency graph

The histogram is a special diagram that is used to represent a frequency

distribution such as that for grouped heights shown above It consists of

a set of rectangles whose areas represent the frequencies of the various

classes Often when producing these diagrams the class width is kept the

same so that the varying frequencies are represented by the height of each

rectangle When drawing histograms for grouped data the midpoints of the

rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc

An adaptation of the histogram known as the frequency polygon may also

be used to represent a frequency distribution

Example 453

Represent the data shown in Table 46 on a histogram and draw in the frequency

polygon for this distribution

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Mathematics for Engineering Technicians 30

All that is required to produce the histogram is to plot frequency against the height

intervals where the intervals are drawn as class widths

Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects

the midpoint of the class widths

Another important method of representation is adding all the frequencies

of a distribution consecutively to produce a graph known as a cumulative

frequency distribution or Ogive

Figure 450 shows the cumulative frequency distribution graph for our data

given in Table 46 while Table 47 shows the consecutive addition of the

frequencies needed to produce the graph in Figure 450

From Figure 450 it is now a simple matter to find for example the median

grouped height or as it is more commonly known the 50th-percentile This

occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this

being in our case 100 giving an equivalent height of approximately 175 cm

Any percentile can be found for example the 75th-percentile where in

our case at a frequency of 150 the height can be seen to be approximately

180 cm

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

0

Class width 5 cm

Height of adults in cm

152 157 162 167 172 177 182 187 192 197

Figure 449 Figure for Example 453 histogram showing frequency distribution

KEY POINTThe frequencies of a distribution may

be added consecutively to produce a

graph known as a cumulative frequency

distribution

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Mathematics for Engineering Technicians308

U N I T 4

C u m u l a t i v e F r e q u e n c y

20

30

40

50

60

70

80

90

100

110

120

130140

150

160

170

180

190

200

10

0

152 157 162 167 172 177 182

180cm

187 192 197

75th percentile

50th percentile

Figure 450 Cumulative frequency distribution graph for data given in Table 46

Table 47 Cumulative frequency data

Height (cm) Frequency Cumulative frequency

150ndash154 4 4

155ndash159 9 13

160ndash164 15 28

165ndash169 21 49

170ndash174 32 81

175ndash179 45 126

180ndash184 41 167

185ndash189 22 189

190ndash194 9 198

195ndash199 2 200

Total 200 200

TYK 411

1 In a particular university the number of students enrolled by a faculty is

given in the table below

Faculty Number of students

Business and administration 1950

Humanities and social science 2820

Physical and life sciences 1050

Technology 850

Total 6670

T e s

t yo u r

k n o

w l e

d g e

TYK

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Mathematics for Engineering Technicians 30

Statistical measurement

When considering statistical data it is often convenient to have one or two

values that represent the data as a whole Average values are often used

You have already found an average value when looking at the median or

50th-percentile of a cumulative frequency distribution So for example we

might talk about the average height of females in the United Kingdom being

170 cm or that the average shoe size of British males is size 9 In statistics

we may represent these average values using the mean median or mode

of the data we are considering We will spend the rest of this short section

finding these average values for both discrete and grouped data starting

with the arithmetic mean

The arithmetic mean

The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the

numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up

and divide by how many there are or more formally

Arithmetic meanarithmetic total of all the individual val

uues

number of values

n

n

sum

where the Greek symbol the sum of the individual values

x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the

data

Illustrate this data on both a bar chart and pie chart

2 For the group of numbers given below produced a tally chart and

determine their frequency of occurrence

36 41 42 38 39 40 42 41 37 40

42 44 43 41 40 38 39 39 43 39

36 37 42 38 39 42 35 42 38 39

40 41 42 37 38 39 44 45 37 40

3 Given the following frequency distribution

Class interval Frequency ( f )

60ndash64 4

65ndash69 11

70ndash74 18

75ndash79 16

80ndash84 7

85ndash90 4

(a) produce a histogram and on it draw the frequency polygon

(b) produce a cumulative frequency graph and from it determine the value

of the 50th-percentile class

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Mathematics for Engineering Technicians310

U N I T 4

So for the mean of our ten numbers we have

mean

n

n

sum 8 7 9 10 5 6 12 9 6 8

10

80

108

Now no matter how long or complex the data we are dealing with provided

that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x

values rsquo is given the symbol x pronounced lsquo x bar rsquo

Example 454

The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm

163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find

the mean height of these females

Then for n 11

x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3

1118483

11168 03

x cm

Mean for grouped data

What if we are required to find the mean for grouped data Look back at

Table 46 showing the height of 200 adults grouped into ten classes In this

case the frequency of the heights needs to be taken into account

We select the class midpoint x as being the average of that class and then

multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the

frequency distribution the total value for the distribution is obtained ( fx )

This total is then divided by the sum of the frequencies ( f ) in order to

determine the mean So for grouped data

x f x f x f x f x

f f f f

f

f

n n

n

1 1 2 2 3 3

1 2 3

sdot

sdot

sumsum

( )midpoint

This rather complicated looking procedure is best illustrated by

example

Example 455

Determine the mean value for the heights of the 200 adults using the data in

Table 46

The values for each individual class are best found by producing a table using

the class midpoints and frequencies and remembering that the class midpoint is found by

dividing the sum of the upper and lower class boundaries by 2 So for example the mean

value for the 1047297rst class interval is149 5 154 5

2152

The completed table is shown

below

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Mathematics for Engineering Technicians 31

Midpoint ( x ) of height (cm) Frequency (f ) fx

152 4 608

157 9 1413

162 15 2430

167 21 3507

172 32 5504

177 45 7965

182 41 7462

187 22 4114

192 9 1728

197 2 394

Total f 200sum fx 35125sum

I hope you can see how each of the values was obtained Now that we have the required

totals the mean value of the distribution can be found

mean value x fx 35125

200175625 05 cm sum

sumf

Notice that our mean value of heights has the same margin of error as the original

measurements The value of the mean cannot be any more accurate than the measured

data from which it was found

Median

When some values within a set of data vary quite widely the arithmetic

mean gives a rather poor representative average of such data Under

these circumstances another more useful measure of the average is the

median

For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which

is not representative of any of the numbers given To find the median value

of the same set of numbers we simply place them in rank order that is 2

3 4 5 6 7 93 Then we select the middle (median) value Since there are

seven numbers (items) we choose the fourth item along the number 5 as

our median value

If the number of items in the set of values is even then we add together the

value of the two middle terms and divide by 2

Example 456

Find the mean and median value for the set of numbers 9 7 8 7 12 70 68

6 5 8

The arithmetic mean is found as

mean x

9 7 8 7 12 70 68 6 5 8

10

200

1020

This value is not really representative of any of the numbers in the set

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Mathematics for Engineering Technicians312

U N I T 4

To 1047297nd the median value we 1047297rst put the numbers in rank order that is

5 6 7 7 8 8 912 68 70

Then from the ten numbers the two middle values The 5th and 6th values along are 8

and 8 So the medianvalue

8 8

2 8

Mode

Yet another measure of central tendency for data containing extreme

values is the mode Now the mode of a set of values containing discrete

data is the value that occurs most often So for the set of values 4 4 4 5

5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs

four times Now it is possible for a set of data to have more than one

mode For example the data used in Example 462 above has two modes

7 and 8 both of these numbers occurring twice and both occurring more

than any of the others A set of data may not have a modal value at all For

example the numbers 2 3 4 5 6 7 8 all occur once and there is

no mode

A set of data that has one mode is called unimodal data with two

modes is bimodal and data with more than two modes is known as

multimodal

When considering frequency distributions for grouped data the modal

class is that group which occurs most frequently If we wish to find

the actual modal value of a frequency distribution we need to draw a

histogram

Example 457

Find the modal class and modal value for the frequency distribution of the height

of adults given in Table 46

Referring back to Table 46 it is easy to see that the class of heights which occurs most

frequently is 175 ndash 179 cm which occurs 45 times

Now to 1047297nd the modal value we need to produce a histogram for the data

We did this for Example 453 This histogram is shown again here with the modalshown

From Figure 451 it can be seen that the modal value 17825 05 cm

This value is obtained from the intersection of the two construction lines AB and CD The

line AB is drawn diagonally from the highest value of the preceding class up to the top

right-hand corner of the modal class The line CD is drawn from the top left-hand corner

of the modal group to the lowest value of the next class immediately above the modal

group Then as can be seen the modal value is read-off where the projection line meets

the x -axis

KEY POINT

The mean median and mode are

statistical averages or measures

of central tendency for a statistical

distribution

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Mathematics for Engineering Technicians 31

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

C

A

B

D

0Height of adults (cm)

152 157 162 167 172 177 182 187 192 197

Modal value 17825 05 cm

Figure 451 Histogram showing frequency distribution and modal value for height of adults

TYK 412

1 Calculate the mean of the numbers 1765 986 1124 1898 959 and

888

2 Determine the mean the median and the mode for the set of numbers 9 8

7 27 16 3 1 9 4 and 116

3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the

arithmetic mean the median and the mode

4 Estimates for the length of wood required for a shelf were as follows

Length (cm) 35 36 37 38 39 40 41 42

Frequency 1 3 4 8 6 5 3 2

Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table

Length (mm) 167 168 169 170 171

Frequency 2 7 20 8 3

6 Calculate the arithmetic mean for the data shown in the table

Length of rivet (mm) 98 99 995 100 1005 101 102

Frequency 3 18 36 62 56 20 5

T e s t yo u

r k

n o

w l e

d g

e

TYK

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Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present

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Mathematics for Engineering Technicians302

U N I T 4

Example 449

Draw a proportionate bar chart for the employment survey shown in Table 44

using the percentage method

For this method all that is required is to 1047297nd the appropriate percentage of the total

(5000) for each category of employment Then choosing a suitable height of column to

represent 100 mark on the appropriate percentage for each of the 10 employment

categories To save space only the 1047297rst 1047297ve categories of employment have been

calculated

1 private business 750

5000100 15

2 public business 900

5000100 18

3

agriculture 200

5000100 4

4 engineering 300

5000100 6

5 transport 425

5000100 8 5

Similarly manufacture 65 leisure industry 14 education 155 health 10

and other categories 25

Figure 446 shows the completed bar chart

Other categories of bar chart include horizontal bar charts where forinstance Figure 444 is turned through 90deg in a clockwise direction One

last type may be used to depict data given in chronological (time) order

Thus for example the horizontal x -axis is used to represent hours days

years etc while the vertical axis shows the variation of the data with time

Example 450

Represent the following data on a chronological bar chart

Year Number employed in general

engineering (thousands)

2003 800

2004 785

2005 690

2006 670

2007 590

Since we have not been asked to represent the data on any speci1047297c bar chart we will use

the simplest involving only the raw data Then the only concern is the scale we should

use for the vertical axis

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Mathematics for Engineering Technicians 30

To present a true representation the scale should start from zero and extend to say800 (Figure 447 a ) If we wish to emphasize a trend that is the way the variable is

rising or falling with time we could use a very much exaggerated scale (Figure 447 b )

This immediately emphasizes the downward trend since 1995 Note that this data is

1047297ctitious (made-up) and used here merely for emphasis

Pie chart

In this type of chart the data is presented as a proportion of the total using

the angle or area of sectors The method used to draw a pie chart is best

illustrated by example

Others (25)

Health (10)

Education (155)

Leisure industry (14)

Manufacture (65)

Transport (85)

Engineering (6)

Agriculture (4)

Public business (18)

Private business (15)

Figure 446 Proportionate percentage bar chart

N u m b e r e m p l o y e

d i n

e n g i n e e r i n g ( t h o u s

a n d s )

1000

900

800

700

600

500

400

300

200

100

02003 2004 2005 2006 2007

Time (years)

(a)

Figure 447 Chronological bar chart (a) in

correct proportion and (b) with graduated

scale

850

800

750

700

650

600

550

500

02003 2004 2005 2006 2007

Time (years)

(b)

N u m b e r e m p l o

y e d i n

e n g i n e e r i n g ( t h o

u s a n d s )

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians304

U N I T 4

Example 451

Represent the data given in Example 450 on a pie chart

Remembering that there are 360deg in a circle and that the total number employed in

general engineering (according to our 1047297gures) was 800 785 690 670 590 3535

(thousands) then we manipulate the data as follows

Year Number employed in general

engineering (thousands)

Sector angle (to nearest half

degree)

2003 800

800

3535360 81 5

2004 785

785

3535360 80

2005 690

690

3535360 70 5

2006 670

670

3535360 68

2007 590

590

3535360 60

Total 3535 360deg

The resulting pie chart is shown in Figure 448

Other methods of visual presentation include pictograms and ideographs

These are diagrams in pictorial form used to present information to thosewho have a limited interest in the subject matter or who do not wish

to deal with data presented in numerical form They have little or no

practical use when interpreting engineering or other scientific data and

apart from acknowledging their existence we will not be pursuing them

further

Frequency distributions

One of the most common and most important ways of organizing and

presenting raw data is through use of frequency distributions

Consider the data given in Table 45 which shows the time in hours that it

took 50 individual workers to complete a specific assembly line task

2004

2005

2006

20072003

Figure 448 Resulting pie chart for

Example 451 employment in engineering

by year

Table 45 Data for assembly line task

11 10 06 11 09 11 08 09 12 07

10 15 09 14 10 09 11 10 10 11

08 09 12 07 06 12 09 08 07 10

10 12 10 10 11 14 07 11 09 09

08 11 10 10 13 05 08 13 13 08

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Mathematics for Engineering Technicians 30

From the data you should be able to see that the shortest time for completion

of the task was 05 hour the longest time was 15 hours The frequency of

appearance of these values is once On the other hand the number of times

the job took 1 hour appears 11 times or it has a frequency of 11 Trying

to sort out the data in this ad hoc manner is time consuming and may lead

to mistakes To assist with the task we use a tally chart This chart simply

shows how many times the event of completing the task in a specific time

takes place To record the frequency of events we use the number 1 in a

tally chart and when the frequency of the event reaches 5 we score through

the existing four 1rsquos to show a frequency of 5 The following example

illustrates the procedure

Example 452

Use a tally chart to determine the frequency of events for the data given on the

assembly line task in Table 45

Time (hours) Tally Frequency

05 1 1

06 11 2

07 1111 4

08 1111 1 6

09 1111 111 8

10 1111 1111 1 11

11 1111 111 8

12 1111 4

13 111 3

14 11 2

15 1 1

Total 50

We now have a full numerical representation of the frequency of events So for example

8 people completed the assembly task in 11 hours or the time 11 hours has a frequency

of 8 We will be using the above information later on when we consider measures of

central tendency

The times in hours given in the above data are simply numbers When data

appears in a form where it can be individually counted we say that it is

discrete data It goes up or down in countable steps Thus the numbers

12 34 86 9 111 130 are said to be discrete If however data is

obtained by measurement for example the heights of a group of people

then we say that this data is continuous When dealing with continuous

data we tend to quote its limits that is the limit of accuracy with which we

take the measurements So for example a person may be 174 05 cm

in height When dealing numerically with continuous data or a large

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians306

U N I T 4

amount of discrete data it is often useful to group this data into classes or

categories We can then find out the numbers (frequency) of items within

each group

Table 46 shows the height of 200 adults grouped into 10 classes

KEY POINT

The grouping of frequency distributions

is a means for clearer presentation of

the facts

Table 46 Height of adults

Height (cm) Frequency

150ndash154 4

155ndash159 9

160ndash164 15

165ndash169 21

170ndash174 32

175ndash179 45

180ndash184 41

185ndash189 22

190ndash194 9

195ndash199 2

Total 200

The main advantage of grouping is that it produces a clear overall picture of

the frequency distribution In Table 46 the first class interval is 150ndash154

The end number 150 is known as the lower limit of the class interval and

the number 154 is the upper limit The heights have been measured to the

nearest centimetre That means within 05 cm Therefore in effect the

first class interval includes all heights in the range 1495ndash1545 cm these

numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and

upper class boundaries not the upper and lower limits of the class interval

The histogram and frequency graph

The histogram is a special diagram that is used to represent a frequency

distribution such as that for grouped heights shown above It consists of

a set of rectangles whose areas represent the frequencies of the various

classes Often when producing these diagrams the class width is kept the

same so that the varying frequencies are represented by the height of each

rectangle When drawing histograms for grouped data the midpoints of the

rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc

An adaptation of the histogram known as the frequency polygon may also

be used to represent a frequency distribution

Example 453

Represent the data shown in Table 46 on a histogram and draw in the frequency

polygon for this distribution

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 30

All that is required to produce the histogram is to plot frequency against the height

intervals where the intervals are drawn as class widths

Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects

the midpoint of the class widths

Another important method of representation is adding all the frequencies

of a distribution consecutively to produce a graph known as a cumulative

frequency distribution or Ogive

Figure 450 shows the cumulative frequency distribution graph for our data

given in Table 46 while Table 47 shows the consecutive addition of the

frequencies needed to produce the graph in Figure 450

From Figure 450 it is now a simple matter to find for example the median

grouped height or as it is more commonly known the 50th-percentile This

occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this

being in our case 100 giving an equivalent height of approximately 175 cm

Any percentile can be found for example the 75th-percentile where in

our case at a frequency of 150 the height can be seen to be approximately

180 cm

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

0

Class width 5 cm

Height of adults in cm

152 157 162 167 172 177 182 187 192 197

Figure 449 Figure for Example 453 histogram showing frequency distribution

KEY POINTThe frequencies of a distribution may

be added consecutively to produce a

graph known as a cumulative frequency

distribution

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Mathematics for Engineering Technicians308

U N I T 4

C u m u l a t i v e F r e q u e n c y

20

30

40

50

60

70

80

90

100

110

120

130140

150

160

170

180

190

200

10

0

152 157 162 167 172 177 182

180cm

187 192 197

75th percentile

50th percentile

Figure 450 Cumulative frequency distribution graph for data given in Table 46

Table 47 Cumulative frequency data

Height (cm) Frequency Cumulative frequency

150ndash154 4 4

155ndash159 9 13

160ndash164 15 28

165ndash169 21 49

170ndash174 32 81

175ndash179 45 126

180ndash184 41 167

185ndash189 22 189

190ndash194 9 198

195ndash199 2 200

Total 200 200

TYK 411

1 In a particular university the number of students enrolled by a faculty is

given in the table below

Faculty Number of students

Business and administration 1950

Humanities and social science 2820

Physical and life sciences 1050

Technology 850

Total 6670

T e s

t yo u r

k n o

w l e

d g e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 30

Statistical measurement

When considering statistical data it is often convenient to have one or two

values that represent the data as a whole Average values are often used

You have already found an average value when looking at the median or

50th-percentile of a cumulative frequency distribution So for example we

might talk about the average height of females in the United Kingdom being

170 cm or that the average shoe size of British males is size 9 In statistics

we may represent these average values using the mean median or mode

of the data we are considering We will spend the rest of this short section

finding these average values for both discrete and grouped data starting

with the arithmetic mean

The arithmetic mean

The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the

numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up

and divide by how many there are or more formally

Arithmetic meanarithmetic total of all the individual val

uues

number of values

n

n

sum

where the Greek symbol the sum of the individual values

x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the

data

Illustrate this data on both a bar chart and pie chart

2 For the group of numbers given below produced a tally chart and

determine their frequency of occurrence

36 41 42 38 39 40 42 41 37 40

42 44 43 41 40 38 39 39 43 39

36 37 42 38 39 42 35 42 38 39

40 41 42 37 38 39 44 45 37 40

3 Given the following frequency distribution

Class interval Frequency ( f )

60ndash64 4

65ndash69 11

70ndash74 18

75ndash79 16

80ndash84 7

85ndash90 4

(a) produce a histogram and on it draw the frequency polygon

(b) produce a cumulative frequency graph and from it determine the value

of the 50th-percentile class

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Mathematics for Engineering Technicians310

U N I T 4

So for the mean of our ten numbers we have

mean

n

n

sum 8 7 9 10 5 6 12 9 6 8

10

80

108

Now no matter how long or complex the data we are dealing with provided

that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x

values rsquo is given the symbol x pronounced lsquo x bar rsquo

Example 454

The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm

163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find

the mean height of these females

Then for n 11

x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3

1118483

11168 03

x cm

Mean for grouped data

What if we are required to find the mean for grouped data Look back at

Table 46 showing the height of 200 adults grouped into ten classes In this

case the frequency of the heights needs to be taken into account

We select the class midpoint x as being the average of that class and then

multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the

frequency distribution the total value for the distribution is obtained ( fx )

This total is then divided by the sum of the frequencies ( f ) in order to

determine the mean So for grouped data

x f x f x f x f x

f f f f

f

f

n n

n

1 1 2 2 3 3

1 2 3

sdot

sdot

sumsum

( )midpoint

This rather complicated looking procedure is best illustrated by

example

Example 455

Determine the mean value for the heights of the 200 adults using the data in

Table 46

The values for each individual class are best found by producing a table using

the class midpoints and frequencies and remembering that the class midpoint is found by

dividing the sum of the upper and lower class boundaries by 2 So for example the mean

value for the 1047297rst class interval is149 5 154 5

2152

The completed table is shown

below

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Mathematics for Engineering Technicians 31

Midpoint ( x ) of height (cm) Frequency (f ) fx

152 4 608

157 9 1413

162 15 2430

167 21 3507

172 32 5504

177 45 7965

182 41 7462

187 22 4114

192 9 1728

197 2 394

Total f 200sum fx 35125sum

I hope you can see how each of the values was obtained Now that we have the required

totals the mean value of the distribution can be found

mean value x fx 35125

200175625 05 cm sum

sumf

Notice that our mean value of heights has the same margin of error as the original

measurements The value of the mean cannot be any more accurate than the measured

data from which it was found

Median

When some values within a set of data vary quite widely the arithmetic

mean gives a rather poor representative average of such data Under

these circumstances another more useful measure of the average is the

median

For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which

is not representative of any of the numbers given To find the median value

of the same set of numbers we simply place them in rank order that is 2

3 4 5 6 7 93 Then we select the middle (median) value Since there are

seven numbers (items) we choose the fourth item along the number 5 as

our median value

If the number of items in the set of values is even then we add together the

value of the two middle terms and divide by 2

Example 456

Find the mean and median value for the set of numbers 9 7 8 7 12 70 68

6 5 8

The arithmetic mean is found as

mean x

9 7 8 7 12 70 68 6 5 8

10

200

1020

This value is not really representative of any of the numbers in the set

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Mathematics for Engineering Technicians312

U N I T 4

To 1047297nd the median value we 1047297rst put the numbers in rank order that is

5 6 7 7 8 8 912 68 70

Then from the ten numbers the two middle values The 5th and 6th values along are 8

and 8 So the medianvalue

8 8

2 8

Mode

Yet another measure of central tendency for data containing extreme

values is the mode Now the mode of a set of values containing discrete

data is the value that occurs most often So for the set of values 4 4 4 5

5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs

four times Now it is possible for a set of data to have more than one

mode For example the data used in Example 462 above has two modes

7 and 8 both of these numbers occurring twice and both occurring more

than any of the others A set of data may not have a modal value at all For

example the numbers 2 3 4 5 6 7 8 all occur once and there is

no mode

A set of data that has one mode is called unimodal data with two

modes is bimodal and data with more than two modes is known as

multimodal

When considering frequency distributions for grouped data the modal

class is that group which occurs most frequently If we wish to find

the actual modal value of a frequency distribution we need to draw a

histogram

Example 457

Find the modal class and modal value for the frequency distribution of the height

of adults given in Table 46

Referring back to Table 46 it is easy to see that the class of heights which occurs most

frequently is 175 ndash 179 cm which occurs 45 times

Now to 1047297nd the modal value we need to produce a histogram for the data

We did this for Example 453 This histogram is shown again here with the modalshown

From Figure 451 it can be seen that the modal value 17825 05 cm

This value is obtained from the intersection of the two construction lines AB and CD The

line AB is drawn diagonally from the highest value of the preceding class up to the top

right-hand corner of the modal class The line CD is drawn from the top left-hand corner

of the modal group to the lowest value of the next class immediately above the modal

group Then as can be seen the modal value is read-off where the projection line meets

the x -axis

KEY POINT

The mean median and mode are

statistical averages or measures

of central tendency for a statistical

distribution

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Mathematics for Engineering Technicians 31

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

C

A

B

D

0Height of adults (cm)

152 157 162 167 172 177 182 187 192 197

Modal value 17825 05 cm

Figure 451 Histogram showing frequency distribution and modal value for height of adults

TYK 412

1 Calculate the mean of the numbers 1765 986 1124 1898 959 and

888

2 Determine the mean the median and the mode for the set of numbers 9 8

7 27 16 3 1 9 4 and 116

3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the

arithmetic mean the median and the mode

4 Estimates for the length of wood required for a shelf were as follows

Length (cm) 35 36 37 38 39 40 41 42

Frequency 1 3 4 8 6 5 3 2

Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table

Length (mm) 167 168 169 170 171

Frequency 2 7 20 8 3

6 Calculate the arithmetic mean for the data shown in the table

Length of rivet (mm) 98 99 995 100 1005 101 102

Frequency 3 18 36 62 56 20 5

T e s t yo u

r k

n o

w l e

d g

e

TYK

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Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present

Page 5: UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 30

To present a true representation the scale should start from zero and extend to say800 (Figure 447 a ) If we wish to emphasize a trend that is the way the variable is

rising or falling with time we could use a very much exaggerated scale (Figure 447 b )

This immediately emphasizes the downward trend since 1995 Note that this data is

1047297ctitious (made-up) and used here merely for emphasis

Pie chart

In this type of chart the data is presented as a proportion of the total using

the angle or area of sectors The method used to draw a pie chart is best

illustrated by example

Others (25)

Health (10)

Education (155)

Leisure industry (14)

Manufacture (65)

Transport (85)

Engineering (6)

Agriculture (4)

Public business (18)

Private business (15)

Figure 446 Proportionate percentage bar chart

N u m b e r e m p l o y e

d i n

e n g i n e e r i n g ( t h o u s

a n d s )

1000

900

800

700

600

500

400

300

200

100

02003 2004 2005 2006 2007

Time (years)

(a)

Figure 447 Chronological bar chart (a) in

correct proportion and (b) with graduated

scale

850

800

750

700

650

600

550

500

02003 2004 2005 2006 2007

Time (years)

(b)

N u m b e r e m p l o

y e d i n

e n g i n e e r i n g ( t h o

u s a n d s )

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians304

U N I T 4

Example 451

Represent the data given in Example 450 on a pie chart

Remembering that there are 360deg in a circle and that the total number employed in

general engineering (according to our 1047297gures) was 800 785 690 670 590 3535

(thousands) then we manipulate the data as follows

Year Number employed in general

engineering (thousands)

Sector angle (to nearest half

degree)

2003 800

800

3535360 81 5

2004 785

785

3535360 80

2005 690

690

3535360 70 5

2006 670

670

3535360 68

2007 590

590

3535360 60

Total 3535 360deg

The resulting pie chart is shown in Figure 448

Other methods of visual presentation include pictograms and ideographs

These are diagrams in pictorial form used to present information to thosewho have a limited interest in the subject matter or who do not wish

to deal with data presented in numerical form They have little or no

practical use when interpreting engineering or other scientific data and

apart from acknowledging their existence we will not be pursuing them

further

Frequency distributions

One of the most common and most important ways of organizing and

presenting raw data is through use of frequency distributions

Consider the data given in Table 45 which shows the time in hours that it

took 50 individual workers to complete a specific assembly line task

2004

2005

2006

20072003

Figure 448 Resulting pie chart for

Example 451 employment in engineering

by year

Table 45 Data for assembly line task

11 10 06 11 09 11 08 09 12 07

10 15 09 14 10 09 11 10 10 11

08 09 12 07 06 12 09 08 07 10

10 12 10 10 11 14 07 11 09 09

08 11 10 10 13 05 08 13 13 08

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Mathematics for Engineering Technicians 30

From the data you should be able to see that the shortest time for completion

of the task was 05 hour the longest time was 15 hours The frequency of

appearance of these values is once On the other hand the number of times

the job took 1 hour appears 11 times or it has a frequency of 11 Trying

to sort out the data in this ad hoc manner is time consuming and may lead

to mistakes To assist with the task we use a tally chart This chart simply

shows how many times the event of completing the task in a specific time

takes place To record the frequency of events we use the number 1 in a

tally chart and when the frequency of the event reaches 5 we score through

the existing four 1rsquos to show a frequency of 5 The following example

illustrates the procedure

Example 452

Use a tally chart to determine the frequency of events for the data given on the

assembly line task in Table 45

Time (hours) Tally Frequency

05 1 1

06 11 2

07 1111 4

08 1111 1 6

09 1111 111 8

10 1111 1111 1 11

11 1111 111 8

12 1111 4

13 111 3

14 11 2

15 1 1

Total 50

We now have a full numerical representation of the frequency of events So for example

8 people completed the assembly task in 11 hours or the time 11 hours has a frequency

of 8 We will be using the above information later on when we consider measures of

central tendency

The times in hours given in the above data are simply numbers When data

appears in a form where it can be individually counted we say that it is

discrete data It goes up or down in countable steps Thus the numbers

12 34 86 9 111 130 are said to be discrete If however data is

obtained by measurement for example the heights of a group of people

then we say that this data is continuous When dealing with continuous

data we tend to quote its limits that is the limit of accuracy with which we

take the measurements So for example a person may be 174 05 cm

in height When dealing numerically with continuous data or a large

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians306

U N I T 4

amount of discrete data it is often useful to group this data into classes or

categories We can then find out the numbers (frequency) of items within

each group

Table 46 shows the height of 200 adults grouped into 10 classes

KEY POINT

The grouping of frequency distributions

is a means for clearer presentation of

the facts

Table 46 Height of adults

Height (cm) Frequency

150ndash154 4

155ndash159 9

160ndash164 15

165ndash169 21

170ndash174 32

175ndash179 45

180ndash184 41

185ndash189 22

190ndash194 9

195ndash199 2

Total 200

The main advantage of grouping is that it produces a clear overall picture of

the frequency distribution In Table 46 the first class interval is 150ndash154

The end number 150 is known as the lower limit of the class interval and

the number 154 is the upper limit The heights have been measured to the

nearest centimetre That means within 05 cm Therefore in effect the

first class interval includes all heights in the range 1495ndash1545 cm these

numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and

upper class boundaries not the upper and lower limits of the class interval

The histogram and frequency graph

The histogram is a special diagram that is used to represent a frequency

distribution such as that for grouped heights shown above It consists of

a set of rectangles whose areas represent the frequencies of the various

classes Often when producing these diagrams the class width is kept the

same so that the varying frequencies are represented by the height of each

rectangle When drawing histograms for grouped data the midpoints of the

rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc

An adaptation of the histogram known as the frequency polygon may also

be used to represent a frequency distribution

Example 453

Represent the data shown in Table 46 on a histogram and draw in the frequency

polygon for this distribution

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Mathematics for Engineering Technicians 30

All that is required to produce the histogram is to plot frequency against the height

intervals where the intervals are drawn as class widths

Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects

the midpoint of the class widths

Another important method of representation is adding all the frequencies

of a distribution consecutively to produce a graph known as a cumulative

frequency distribution or Ogive

Figure 450 shows the cumulative frequency distribution graph for our data

given in Table 46 while Table 47 shows the consecutive addition of the

frequencies needed to produce the graph in Figure 450

From Figure 450 it is now a simple matter to find for example the median

grouped height or as it is more commonly known the 50th-percentile This

occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this

being in our case 100 giving an equivalent height of approximately 175 cm

Any percentile can be found for example the 75th-percentile where in

our case at a frequency of 150 the height can be seen to be approximately

180 cm

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

0

Class width 5 cm

Height of adults in cm

152 157 162 167 172 177 182 187 192 197

Figure 449 Figure for Example 453 histogram showing frequency distribution

KEY POINTThe frequencies of a distribution may

be added consecutively to produce a

graph known as a cumulative frequency

distribution

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians308

U N I T 4

C u m u l a t i v e F r e q u e n c y

20

30

40

50

60

70

80

90

100

110

120

130140

150

160

170

180

190

200

10

0

152 157 162 167 172 177 182

180cm

187 192 197

75th percentile

50th percentile

Figure 450 Cumulative frequency distribution graph for data given in Table 46

Table 47 Cumulative frequency data

Height (cm) Frequency Cumulative frequency

150ndash154 4 4

155ndash159 9 13

160ndash164 15 28

165ndash169 21 49

170ndash174 32 81

175ndash179 45 126

180ndash184 41 167

185ndash189 22 189

190ndash194 9 198

195ndash199 2 200

Total 200 200

TYK 411

1 In a particular university the number of students enrolled by a faculty is

given in the table below

Faculty Number of students

Business and administration 1950

Humanities and social science 2820

Physical and life sciences 1050

Technology 850

Total 6670

T e s

t yo u r

k n o

w l e

d g e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1116

Mathematics for Engineering Technicians 30

Statistical measurement

When considering statistical data it is often convenient to have one or two

values that represent the data as a whole Average values are often used

You have already found an average value when looking at the median or

50th-percentile of a cumulative frequency distribution So for example we

might talk about the average height of females in the United Kingdom being

170 cm or that the average shoe size of British males is size 9 In statistics

we may represent these average values using the mean median or mode

of the data we are considering We will spend the rest of this short section

finding these average values for both discrete and grouped data starting

with the arithmetic mean

The arithmetic mean

The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the

numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up

and divide by how many there are or more formally

Arithmetic meanarithmetic total of all the individual val

uues

number of values

n

n

sum

where the Greek symbol the sum of the individual values

x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the

data

Illustrate this data on both a bar chart and pie chart

2 For the group of numbers given below produced a tally chart and

determine their frequency of occurrence

36 41 42 38 39 40 42 41 37 40

42 44 43 41 40 38 39 39 43 39

36 37 42 38 39 42 35 42 38 39

40 41 42 37 38 39 44 45 37 40

3 Given the following frequency distribution

Class interval Frequency ( f )

60ndash64 4

65ndash69 11

70ndash74 18

75ndash79 16

80ndash84 7

85ndash90 4

(a) produce a histogram and on it draw the frequency polygon

(b) produce a cumulative frequency graph and from it determine the value

of the 50th-percentile class

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians310

U N I T 4

So for the mean of our ten numbers we have

mean

n

n

sum 8 7 9 10 5 6 12 9 6 8

10

80

108

Now no matter how long or complex the data we are dealing with provided

that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x

values rsquo is given the symbol x pronounced lsquo x bar rsquo

Example 454

The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm

163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find

the mean height of these females

Then for n 11

x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3

1118483

11168 03

x cm

Mean for grouped data

What if we are required to find the mean for grouped data Look back at

Table 46 showing the height of 200 adults grouped into ten classes In this

case the frequency of the heights needs to be taken into account

We select the class midpoint x as being the average of that class and then

multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the

frequency distribution the total value for the distribution is obtained ( fx )

This total is then divided by the sum of the frequencies ( f ) in order to

determine the mean So for grouped data

x f x f x f x f x

f f f f

f

f

n n

n

1 1 2 2 3 3

1 2 3

sdot

sdot

sumsum

( )midpoint

This rather complicated looking procedure is best illustrated by

example

Example 455

Determine the mean value for the heights of the 200 adults using the data in

Table 46

The values for each individual class are best found by producing a table using

the class midpoints and frequencies and remembering that the class midpoint is found by

dividing the sum of the upper and lower class boundaries by 2 So for example the mean

value for the 1047297rst class interval is149 5 154 5

2152

The completed table is shown

below

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Mathematics for Engineering Technicians 31

Midpoint ( x ) of height (cm) Frequency (f ) fx

152 4 608

157 9 1413

162 15 2430

167 21 3507

172 32 5504

177 45 7965

182 41 7462

187 22 4114

192 9 1728

197 2 394

Total f 200sum fx 35125sum

I hope you can see how each of the values was obtained Now that we have the required

totals the mean value of the distribution can be found

mean value x fx 35125

200175625 05 cm sum

sumf

Notice that our mean value of heights has the same margin of error as the original

measurements The value of the mean cannot be any more accurate than the measured

data from which it was found

Median

When some values within a set of data vary quite widely the arithmetic

mean gives a rather poor representative average of such data Under

these circumstances another more useful measure of the average is the

median

For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which

is not representative of any of the numbers given To find the median value

of the same set of numbers we simply place them in rank order that is 2

3 4 5 6 7 93 Then we select the middle (median) value Since there are

seven numbers (items) we choose the fourth item along the number 5 as

our median value

If the number of items in the set of values is even then we add together the

value of the two middle terms and divide by 2

Example 456

Find the mean and median value for the set of numbers 9 7 8 7 12 70 68

6 5 8

The arithmetic mean is found as

mean x

9 7 8 7 12 70 68 6 5 8

10

200

1020

This value is not really representative of any of the numbers in the set

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians312

U N I T 4

To 1047297nd the median value we 1047297rst put the numbers in rank order that is

5 6 7 7 8 8 912 68 70

Then from the ten numbers the two middle values The 5th and 6th values along are 8

and 8 So the medianvalue

8 8

2 8

Mode

Yet another measure of central tendency for data containing extreme

values is the mode Now the mode of a set of values containing discrete

data is the value that occurs most often So for the set of values 4 4 4 5

5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs

four times Now it is possible for a set of data to have more than one

mode For example the data used in Example 462 above has two modes

7 and 8 both of these numbers occurring twice and both occurring more

than any of the others A set of data may not have a modal value at all For

example the numbers 2 3 4 5 6 7 8 all occur once and there is

no mode

A set of data that has one mode is called unimodal data with two

modes is bimodal and data with more than two modes is known as

multimodal

When considering frequency distributions for grouped data the modal

class is that group which occurs most frequently If we wish to find

the actual modal value of a frequency distribution we need to draw a

histogram

Example 457

Find the modal class and modal value for the frequency distribution of the height

of adults given in Table 46

Referring back to Table 46 it is easy to see that the class of heights which occurs most

frequently is 175 ndash 179 cm which occurs 45 times

Now to 1047297nd the modal value we need to produce a histogram for the data

We did this for Example 453 This histogram is shown again here with the modalshown

From Figure 451 it can be seen that the modal value 17825 05 cm

This value is obtained from the intersection of the two construction lines AB and CD The

line AB is drawn diagonally from the highest value of the preceding class up to the top

right-hand corner of the modal class The line CD is drawn from the top left-hand corner

of the modal group to the lowest value of the next class immediately above the modal

group Then as can be seen the modal value is read-off where the projection line meets

the x -axis

KEY POINT

The mean median and mode are

statistical averages or measures

of central tendency for a statistical

distribution

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Mathematics for Engineering Technicians 31

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

C

A

B

D

0Height of adults (cm)

152 157 162 167 172 177 182 187 192 197

Modal value 17825 05 cm

Figure 451 Histogram showing frequency distribution and modal value for height of adults

TYK 412

1 Calculate the mean of the numbers 1765 986 1124 1898 959 and

888

2 Determine the mean the median and the mode for the set of numbers 9 8

7 27 16 3 1 9 4 and 116

3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the

arithmetic mean the median and the mode

4 Estimates for the length of wood required for a shelf were as follows

Length (cm) 35 36 37 38 39 40 41 42

Frequency 1 3 4 8 6 5 3 2

Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table

Length (mm) 167 168 169 170 171

Frequency 2 7 20 8 3

6 Calculate the arithmetic mean for the data shown in the table

Length of rivet (mm) 98 99 995 100 1005 101 102

Frequency 3 18 36 62 56 20 5

T e s t yo u

r k

n o

w l e

d g

e

TYK

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Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present

Page 6: UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians304

U N I T 4

Example 451

Represent the data given in Example 450 on a pie chart

Remembering that there are 360deg in a circle and that the total number employed in

general engineering (according to our 1047297gures) was 800 785 690 670 590 3535

(thousands) then we manipulate the data as follows

Year Number employed in general

engineering (thousands)

Sector angle (to nearest half

degree)

2003 800

800

3535360 81 5

2004 785

785

3535360 80

2005 690

690

3535360 70 5

2006 670

670

3535360 68

2007 590

590

3535360 60

Total 3535 360deg

The resulting pie chart is shown in Figure 448

Other methods of visual presentation include pictograms and ideographs

These are diagrams in pictorial form used to present information to thosewho have a limited interest in the subject matter or who do not wish

to deal with data presented in numerical form They have little or no

practical use when interpreting engineering or other scientific data and

apart from acknowledging their existence we will not be pursuing them

further

Frequency distributions

One of the most common and most important ways of organizing and

presenting raw data is through use of frequency distributions

Consider the data given in Table 45 which shows the time in hours that it

took 50 individual workers to complete a specific assembly line task

2004

2005

2006

20072003

Figure 448 Resulting pie chart for

Example 451 employment in engineering

by year

Table 45 Data for assembly line task

11 10 06 11 09 11 08 09 12 07

10 15 09 14 10 09 11 10 10 11

08 09 12 07 06 12 09 08 07 10

10 12 10 10 11 14 07 11 09 09

08 11 10 10 13 05 08 13 13 08

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Mathematics for Engineering Technicians 30

From the data you should be able to see that the shortest time for completion

of the task was 05 hour the longest time was 15 hours The frequency of

appearance of these values is once On the other hand the number of times

the job took 1 hour appears 11 times or it has a frequency of 11 Trying

to sort out the data in this ad hoc manner is time consuming and may lead

to mistakes To assist with the task we use a tally chart This chart simply

shows how many times the event of completing the task in a specific time

takes place To record the frequency of events we use the number 1 in a

tally chart and when the frequency of the event reaches 5 we score through

the existing four 1rsquos to show a frequency of 5 The following example

illustrates the procedure

Example 452

Use a tally chart to determine the frequency of events for the data given on the

assembly line task in Table 45

Time (hours) Tally Frequency

05 1 1

06 11 2

07 1111 4

08 1111 1 6

09 1111 111 8

10 1111 1111 1 11

11 1111 111 8

12 1111 4

13 111 3

14 11 2

15 1 1

Total 50

We now have a full numerical representation of the frequency of events So for example

8 people completed the assembly task in 11 hours or the time 11 hours has a frequency

of 8 We will be using the above information later on when we consider measures of

central tendency

The times in hours given in the above data are simply numbers When data

appears in a form where it can be individually counted we say that it is

discrete data It goes up or down in countable steps Thus the numbers

12 34 86 9 111 130 are said to be discrete If however data is

obtained by measurement for example the heights of a group of people

then we say that this data is continuous When dealing with continuous

data we tend to quote its limits that is the limit of accuracy with which we

take the measurements So for example a person may be 174 05 cm

in height When dealing numerically with continuous data or a large

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Mathematics for Engineering Technicians306

U N I T 4

amount of discrete data it is often useful to group this data into classes or

categories We can then find out the numbers (frequency) of items within

each group

Table 46 shows the height of 200 adults grouped into 10 classes

KEY POINT

The grouping of frequency distributions

is a means for clearer presentation of

the facts

Table 46 Height of adults

Height (cm) Frequency

150ndash154 4

155ndash159 9

160ndash164 15

165ndash169 21

170ndash174 32

175ndash179 45

180ndash184 41

185ndash189 22

190ndash194 9

195ndash199 2

Total 200

The main advantage of grouping is that it produces a clear overall picture of

the frequency distribution In Table 46 the first class interval is 150ndash154

The end number 150 is known as the lower limit of the class interval and

the number 154 is the upper limit The heights have been measured to the

nearest centimetre That means within 05 cm Therefore in effect the

first class interval includes all heights in the range 1495ndash1545 cm these

numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and

upper class boundaries not the upper and lower limits of the class interval

The histogram and frequency graph

The histogram is a special diagram that is used to represent a frequency

distribution such as that for grouped heights shown above It consists of

a set of rectangles whose areas represent the frequencies of the various

classes Often when producing these diagrams the class width is kept the

same so that the varying frequencies are represented by the height of each

rectangle When drawing histograms for grouped data the midpoints of the

rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc

An adaptation of the histogram known as the frequency polygon may also

be used to represent a frequency distribution

Example 453

Represent the data shown in Table 46 on a histogram and draw in the frequency

polygon for this distribution

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Mathematics for Engineering Technicians 30

All that is required to produce the histogram is to plot frequency against the height

intervals where the intervals are drawn as class widths

Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects

the midpoint of the class widths

Another important method of representation is adding all the frequencies

of a distribution consecutively to produce a graph known as a cumulative

frequency distribution or Ogive

Figure 450 shows the cumulative frequency distribution graph for our data

given in Table 46 while Table 47 shows the consecutive addition of the

frequencies needed to produce the graph in Figure 450

From Figure 450 it is now a simple matter to find for example the median

grouped height or as it is more commonly known the 50th-percentile This

occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this

being in our case 100 giving an equivalent height of approximately 175 cm

Any percentile can be found for example the 75th-percentile where in

our case at a frequency of 150 the height can be seen to be approximately

180 cm

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

0

Class width 5 cm

Height of adults in cm

152 157 162 167 172 177 182 187 192 197

Figure 449 Figure for Example 453 histogram showing frequency distribution

KEY POINTThe frequencies of a distribution may

be added consecutively to produce a

graph known as a cumulative frequency

distribution

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians308

U N I T 4

C u m u l a t i v e F r e q u e n c y

20

30

40

50

60

70

80

90

100

110

120

130140

150

160

170

180

190

200

10

0

152 157 162 167 172 177 182

180cm

187 192 197

75th percentile

50th percentile

Figure 450 Cumulative frequency distribution graph for data given in Table 46

Table 47 Cumulative frequency data

Height (cm) Frequency Cumulative frequency

150ndash154 4 4

155ndash159 9 13

160ndash164 15 28

165ndash169 21 49

170ndash174 32 81

175ndash179 45 126

180ndash184 41 167

185ndash189 22 189

190ndash194 9 198

195ndash199 2 200

Total 200 200

TYK 411

1 In a particular university the number of students enrolled by a faculty is

given in the table below

Faculty Number of students

Business and administration 1950

Humanities and social science 2820

Physical and life sciences 1050

Technology 850

Total 6670

T e s

t yo u r

k n o

w l e

d g e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 30

Statistical measurement

When considering statistical data it is often convenient to have one or two

values that represent the data as a whole Average values are often used

You have already found an average value when looking at the median or

50th-percentile of a cumulative frequency distribution So for example we

might talk about the average height of females in the United Kingdom being

170 cm or that the average shoe size of British males is size 9 In statistics

we may represent these average values using the mean median or mode

of the data we are considering We will spend the rest of this short section

finding these average values for both discrete and grouped data starting

with the arithmetic mean

The arithmetic mean

The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the

numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up

and divide by how many there are or more formally

Arithmetic meanarithmetic total of all the individual val

uues

number of values

n

n

sum

where the Greek symbol the sum of the individual values

x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the

data

Illustrate this data on both a bar chart and pie chart

2 For the group of numbers given below produced a tally chart and

determine their frequency of occurrence

36 41 42 38 39 40 42 41 37 40

42 44 43 41 40 38 39 39 43 39

36 37 42 38 39 42 35 42 38 39

40 41 42 37 38 39 44 45 37 40

3 Given the following frequency distribution

Class interval Frequency ( f )

60ndash64 4

65ndash69 11

70ndash74 18

75ndash79 16

80ndash84 7

85ndash90 4

(a) produce a histogram and on it draw the frequency polygon

(b) produce a cumulative frequency graph and from it determine the value

of the 50th-percentile class

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians310

U N I T 4

So for the mean of our ten numbers we have

mean

n

n

sum 8 7 9 10 5 6 12 9 6 8

10

80

108

Now no matter how long or complex the data we are dealing with provided

that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x

values rsquo is given the symbol x pronounced lsquo x bar rsquo

Example 454

The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm

163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find

the mean height of these females

Then for n 11

x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3

1118483

11168 03

x cm

Mean for grouped data

What if we are required to find the mean for grouped data Look back at

Table 46 showing the height of 200 adults grouped into ten classes In this

case the frequency of the heights needs to be taken into account

We select the class midpoint x as being the average of that class and then

multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the

frequency distribution the total value for the distribution is obtained ( fx )

This total is then divided by the sum of the frequencies ( f ) in order to

determine the mean So for grouped data

x f x f x f x f x

f f f f

f

f

n n

n

1 1 2 2 3 3

1 2 3

sdot

sdot

sumsum

( )midpoint

This rather complicated looking procedure is best illustrated by

example

Example 455

Determine the mean value for the heights of the 200 adults using the data in

Table 46

The values for each individual class are best found by producing a table using

the class midpoints and frequencies and remembering that the class midpoint is found by

dividing the sum of the upper and lower class boundaries by 2 So for example the mean

value for the 1047297rst class interval is149 5 154 5

2152

The completed table is shown

below

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 31

Midpoint ( x ) of height (cm) Frequency (f ) fx

152 4 608

157 9 1413

162 15 2430

167 21 3507

172 32 5504

177 45 7965

182 41 7462

187 22 4114

192 9 1728

197 2 394

Total f 200sum fx 35125sum

I hope you can see how each of the values was obtained Now that we have the required

totals the mean value of the distribution can be found

mean value x fx 35125

200175625 05 cm sum

sumf

Notice that our mean value of heights has the same margin of error as the original

measurements The value of the mean cannot be any more accurate than the measured

data from which it was found

Median

When some values within a set of data vary quite widely the arithmetic

mean gives a rather poor representative average of such data Under

these circumstances another more useful measure of the average is the

median

For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which

is not representative of any of the numbers given To find the median value

of the same set of numbers we simply place them in rank order that is 2

3 4 5 6 7 93 Then we select the middle (median) value Since there are

seven numbers (items) we choose the fourth item along the number 5 as

our median value

If the number of items in the set of values is even then we add together the

value of the two middle terms and divide by 2

Example 456

Find the mean and median value for the set of numbers 9 7 8 7 12 70 68

6 5 8

The arithmetic mean is found as

mean x

9 7 8 7 12 70 68 6 5 8

10

200

1020

This value is not really representative of any of the numbers in the set

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Mathematics for Engineering Technicians312

U N I T 4

To 1047297nd the median value we 1047297rst put the numbers in rank order that is

5 6 7 7 8 8 912 68 70

Then from the ten numbers the two middle values The 5th and 6th values along are 8

and 8 So the medianvalue

8 8

2 8

Mode

Yet another measure of central tendency for data containing extreme

values is the mode Now the mode of a set of values containing discrete

data is the value that occurs most often So for the set of values 4 4 4 5

5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs

four times Now it is possible for a set of data to have more than one

mode For example the data used in Example 462 above has two modes

7 and 8 both of these numbers occurring twice and both occurring more

than any of the others A set of data may not have a modal value at all For

example the numbers 2 3 4 5 6 7 8 all occur once and there is

no mode

A set of data that has one mode is called unimodal data with two

modes is bimodal and data with more than two modes is known as

multimodal

When considering frequency distributions for grouped data the modal

class is that group which occurs most frequently If we wish to find

the actual modal value of a frequency distribution we need to draw a

histogram

Example 457

Find the modal class and modal value for the frequency distribution of the height

of adults given in Table 46

Referring back to Table 46 it is easy to see that the class of heights which occurs most

frequently is 175 ndash 179 cm which occurs 45 times

Now to 1047297nd the modal value we need to produce a histogram for the data

We did this for Example 453 This histogram is shown again here with the modalshown

From Figure 451 it can be seen that the modal value 17825 05 cm

This value is obtained from the intersection of the two construction lines AB and CD The

line AB is drawn diagonally from the highest value of the preceding class up to the top

right-hand corner of the modal class The line CD is drawn from the top left-hand corner

of the modal group to the lowest value of the next class immediately above the modal

group Then as can be seen the modal value is read-off where the projection line meets

the x -axis

KEY POINT

The mean median and mode are

statistical averages or measures

of central tendency for a statistical

distribution

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Mathematics for Engineering Technicians 31

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

C

A

B

D

0Height of adults (cm)

152 157 162 167 172 177 182 187 192 197

Modal value 17825 05 cm

Figure 451 Histogram showing frequency distribution and modal value for height of adults

TYK 412

1 Calculate the mean of the numbers 1765 986 1124 1898 959 and

888

2 Determine the mean the median and the mode for the set of numbers 9 8

7 27 16 3 1 9 4 and 116

3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the

arithmetic mean the median and the mode

4 Estimates for the length of wood required for a shelf were as follows

Length (cm) 35 36 37 38 39 40 41 42

Frequency 1 3 4 8 6 5 3 2

Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table

Length (mm) 167 168 169 170 171

Frequency 2 7 20 8 3

6 Calculate the arithmetic mean for the data shown in the table

Length of rivet (mm) 98 99 995 100 1005 101 102

Frequency 3 18 36 62 56 20 5

T e s t yo u

r k

n o

w l e

d g

e

TYK

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Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present

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Mathematics for Engineering Technicians 30

From the data you should be able to see that the shortest time for completion

of the task was 05 hour the longest time was 15 hours The frequency of

appearance of these values is once On the other hand the number of times

the job took 1 hour appears 11 times or it has a frequency of 11 Trying

to sort out the data in this ad hoc manner is time consuming and may lead

to mistakes To assist with the task we use a tally chart This chart simply

shows how many times the event of completing the task in a specific time

takes place To record the frequency of events we use the number 1 in a

tally chart and when the frequency of the event reaches 5 we score through

the existing four 1rsquos to show a frequency of 5 The following example

illustrates the procedure

Example 452

Use a tally chart to determine the frequency of events for the data given on the

assembly line task in Table 45

Time (hours) Tally Frequency

05 1 1

06 11 2

07 1111 4

08 1111 1 6

09 1111 111 8

10 1111 1111 1 11

11 1111 111 8

12 1111 4

13 111 3

14 11 2

15 1 1

Total 50

We now have a full numerical representation of the frequency of events So for example

8 people completed the assembly task in 11 hours or the time 11 hours has a frequency

of 8 We will be using the above information later on when we consider measures of

central tendency

The times in hours given in the above data are simply numbers When data

appears in a form where it can be individually counted we say that it is

discrete data It goes up or down in countable steps Thus the numbers

12 34 86 9 111 130 are said to be discrete If however data is

obtained by measurement for example the heights of a group of people

then we say that this data is continuous When dealing with continuous

data we tend to quote its limits that is the limit of accuracy with which we

take the measurements So for example a person may be 174 05 cm

in height When dealing numerically with continuous data or a large

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Mathematics for Engineering Technicians306

U N I T 4

amount of discrete data it is often useful to group this data into classes or

categories We can then find out the numbers (frequency) of items within

each group

Table 46 shows the height of 200 adults grouped into 10 classes

KEY POINT

The grouping of frequency distributions

is a means for clearer presentation of

the facts

Table 46 Height of adults

Height (cm) Frequency

150ndash154 4

155ndash159 9

160ndash164 15

165ndash169 21

170ndash174 32

175ndash179 45

180ndash184 41

185ndash189 22

190ndash194 9

195ndash199 2

Total 200

The main advantage of grouping is that it produces a clear overall picture of

the frequency distribution In Table 46 the first class interval is 150ndash154

The end number 150 is known as the lower limit of the class interval and

the number 154 is the upper limit The heights have been measured to the

nearest centimetre That means within 05 cm Therefore in effect the

first class interval includes all heights in the range 1495ndash1545 cm these

numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and

upper class boundaries not the upper and lower limits of the class interval

The histogram and frequency graph

The histogram is a special diagram that is used to represent a frequency

distribution such as that for grouped heights shown above It consists of

a set of rectangles whose areas represent the frequencies of the various

classes Often when producing these diagrams the class width is kept the

same so that the varying frequencies are represented by the height of each

rectangle When drawing histograms for grouped data the midpoints of the

rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc

An adaptation of the histogram known as the frequency polygon may also

be used to represent a frequency distribution

Example 453

Represent the data shown in Table 46 on a histogram and draw in the frequency

polygon for this distribution

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Mathematics for Engineering Technicians 30

All that is required to produce the histogram is to plot frequency against the height

intervals where the intervals are drawn as class widths

Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects

the midpoint of the class widths

Another important method of representation is adding all the frequencies

of a distribution consecutively to produce a graph known as a cumulative

frequency distribution or Ogive

Figure 450 shows the cumulative frequency distribution graph for our data

given in Table 46 while Table 47 shows the consecutive addition of the

frequencies needed to produce the graph in Figure 450

From Figure 450 it is now a simple matter to find for example the median

grouped height or as it is more commonly known the 50th-percentile This

occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this

being in our case 100 giving an equivalent height of approximately 175 cm

Any percentile can be found for example the 75th-percentile where in

our case at a frequency of 150 the height can be seen to be approximately

180 cm

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

0

Class width 5 cm

Height of adults in cm

152 157 162 167 172 177 182 187 192 197

Figure 449 Figure for Example 453 histogram showing frequency distribution

KEY POINTThe frequencies of a distribution may

be added consecutively to produce a

graph known as a cumulative frequency

distribution

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Mathematics for Engineering Technicians308

U N I T 4

C u m u l a t i v e F r e q u e n c y

20

30

40

50

60

70

80

90

100

110

120

130140

150

160

170

180

190

200

10

0

152 157 162 167 172 177 182

180cm

187 192 197

75th percentile

50th percentile

Figure 450 Cumulative frequency distribution graph for data given in Table 46

Table 47 Cumulative frequency data

Height (cm) Frequency Cumulative frequency

150ndash154 4 4

155ndash159 9 13

160ndash164 15 28

165ndash169 21 49

170ndash174 32 81

175ndash179 45 126

180ndash184 41 167

185ndash189 22 189

190ndash194 9 198

195ndash199 2 200

Total 200 200

TYK 411

1 In a particular university the number of students enrolled by a faculty is

given in the table below

Faculty Number of students

Business and administration 1950

Humanities and social science 2820

Physical and life sciences 1050

Technology 850

Total 6670

T e s

t yo u r

k n o

w l e

d g e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 30

Statistical measurement

When considering statistical data it is often convenient to have one or two

values that represent the data as a whole Average values are often used

You have already found an average value when looking at the median or

50th-percentile of a cumulative frequency distribution So for example we

might talk about the average height of females in the United Kingdom being

170 cm or that the average shoe size of British males is size 9 In statistics

we may represent these average values using the mean median or mode

of the data we are considering We will spend the rest of this short section

finding these average values for both discrete and grouped data starting

with the arithmetic mean

The arithmetic mean

The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the

numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up

and divide by how many there are or more formally

Arithmetic meanarithmetic total of all the individual val

uues

number of values

n

n

sum

where the Greek symbol the sum of the individual values

x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the

data

Illustrate this data on both a bar chart and pie chart

2 For the group of numbers given below produced a tally chart and

determine their frequency of occurrence

36 41 42 38 39 40 42 41 37 40

42 44 43 41 40 38 39 39 43 39

36 37 42 38 39 42 35 42 38 39

40 41 42 37 38 39 44 45 37 40

3 Given the following frequency distribution

Class interval Frequency ( f )

60ndash64 4

65ndash69 11

70ndash74 18

75ndash79 16

80ndash84 7

85ndash90 4

(a) produce a histogram and on it draw the frequency polygon

(b) produce a cumulative frequency graph and from it determine the value

of the 50th-percentile class

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Mathematics for Engineering Technicians310

U N I T 4

So for the mean of our ten numbers we have

mean

n

n

sum 8 7 9 10 5 6 12 9 6 8

10

80

108

Now no matter how long or complex the data we are dealing with provided

that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x

values rsquo is given the symbol x pronounced lsquo x bar rsquo

Example 454

The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm

163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find

the mean height of these females

Then for n 11

x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3

1118483

11168 03

x cm

Mean for grouped data

What if we are required to find the mean for grouped data Look back at

Table 46 showing the height of 200 adults grouped into ten classes In this

case the frequency of the heights needs to be taken into account

We select the class midpoint x as being the average of that class and then

multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the

frequency distribution the total value for the distribution is obtained ( fx )

This total is then divided by the sum of the frequencies ( f ) in order to

determine the mean So for grouped data

x f x f x f x f x

f f f f

f

f

n n

n

1 1 2 2 3 3

1 2 3

sdot

sdot

sumsum

( )midpoint

This rather complicated looking procedure is best illustrated by

example

Example 455

Determine the mean value for the heights of the 200 adults using the data in

Table 46

The values for each individual class are best found by producing a table using

the class midpoints and frequencies and remembering that the class midpoint is found by

dividing the sum of the upper and lower class boundaries by 2 So for example the mean

value for the 1047297rst class interval is149 5 154 5

2152

The completed table is shown

below

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Mathematics for Engineering Technicians 31

Midpoint ( x ) of height (cm) Frequency (f ) fx

152 4 608

157 9 1413

162 15 2430

167 21 3507

172 32 5504

177 45 7965

182 41 7462

187 22 4114

192 9 1728

197 2 394

Total f 200sum fx 35125sum

I hope you can see how each of the values was obtained Now that we have the required

totals the mean value of the distribution can be found

mean value x fx 35125

200175625 05 cm sum

sumf

Notice that our mean value of heights has the same margin of error as the original

measurements The value of the mean cannot be any more accurate than the measured

data from which it was found

Median

When some values within a set of data vary quite widely the arithmetic

mean gives a rather poor representative average of such data Under

these circumstances another more useful measure of the average is the

median

For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which

is not representative of any of the numbers given To find the median value

of the same set of numbers we simply place them in rank order that is 2

3 4 5 6 7 93 Then we select the middle (median) value Since there are

seven numbers (items) we choose the fourth item along the number 5 as

our median value

If the number of items in the set of values is even then we add together the

value of the two middle terms and divide by 2

Example 456

Find the mean and median value for the set of numbers 9 7 8 7 12 70 68

6 5 8

The arithmetic mean is found as

mean x

9 7 8 7 12 70 68 6 5 8

10

200

1020

This value is not really representative of any of the numbers in the set

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Mathematics for Engineering Technicians312

U N I T 4

To 1047297nd the median value we 1047297rst put the numbers in rank order that is

5 6 7 7 8 8 912 68 70

Then from the ten numbers the two middle values The 5th and 6th values along are 8

and 8 So the medianvalue

8 8

2 8

Mode

Yet another measure of central tendency for data containing extreme

values is the mode Now the mode of a set of values containing discrete

data is the value that occurs most often So for the set of values 4 4 4 5

5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs

four times Now it is possible for a set of data to have more than one

mode For example the data used in Example 462 above has two modes

7 and 8 both of these numbers occurring twice and both occurring more

than any of the others A set of data may not have a modal value at all For

example the numbers 2 3 4 5 6 7 8 all occur once and there is

no mode

A set of data that has one mode is called unimodal data with two

modes is bimodal and data with more than two modes is known as

multimodal

When considering frequency distributions for grouped data the modal

class is that group which occurs most frequently If we wish to find

the actual modal value of a frequency distribution we need to draw a

histogram

Example 457

Find the modal class and modal value for the frequency distribution of the height

of adults given in Table 46

Referring back to Table 46 it is easy to see that the class of heights which occurs most

frequently is 175 ndash 179 cm which occurs 45 times

Now to 1047297nd the modal value we need to produce a histogram for the data

We did this for Example 453 This histogram is shown again here with the modalshown

From Figure 451 it can be seen that the modal value 17825 05 cm

This value is obtained from the intersection of the two construction lines AB and CD The

line AB is drawn diagonally from the highest value of the preceding class up to the top

right-hand corner of the modal class The line CD is drawn from the top left-hand corner

of the modal group to the lowest value of the next class immediately above the modal

group Then as can be seen the modal value is read-off where the projection line meets

the x -axis

KEY POINT

The mean median and mode are

statistical averages or measures

of central tendency for a statistical

distribution

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Mathematics for Engineering Technicians 31

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

C

A

B

D

0Height of adults (cm)

152 157 162 167 172 177 182 187 192 197

Modal value 17825 05 cm

Figure 451 Histogram showing frequency distribution and modal value for height of adults

TYK 412

1 Calculate the mean of the numbers 1765 986 1124 1898 959 and

888

2 Determine the mean the median and the mode for the set of numbers 9 8

7 27 16 3 1 9 4 and 116

3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the

arithmetic mean the median and the mode

4 Estimates for the length of wood required for a shelf were as follows

Length (cm) 35 36 37 38 39 40 41 42

Frequency 1 3 4 8 6 5 3 2

Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table

Length (mm) 167 168 169 170 171

Frequency 2 7 20 8 3

6 Calculate the arithmetic mean for the data shown in the table

Length of rivet (mm) 98 99 995 100 1005 101 102

Frequency 3 18 36 62 56 20 5

T e s t yo u

r k

n o

w l e

d g

e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present

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Mathematics for Engineering Technicians306

U N I T 4

amount of discrete data it is often useful to group this data into classes or

categories We can then find out the numbers (frequency) of items within

each group

Table 46 shows the height of 200 adults grouped into 10 classes

KEY POINT

The grouping of frequency distributions

is a means for clearer presentation of

the facts

Table 46 Height of adults

Height (cm) Frequency

150ndash154 4

155ndash159 9

160ndash164 15

165ndash169 21

170ndash174 32

175ndash179 45

180ndash184 41

185ndash189 22

190ndash194 9

195ndash199 2

Total 200

The main advantage of grouping is that it produces a clear overall picture of

the frequency distribution In Table 46 the first class interval is 150ndash154

The end number 150 is known as the lower limit of the class interval and

the number 154 is the upper limit The heights have been measured to the

nearest centimetre That means within 05 cm Therefore in effect the

first class interval includes all heights in the range 1495ndash1545 cm these

numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and

upper class boundaries not the upper and lower limits of the class interval

The histogram and frequency graph

The histogram is a special diagram that is used to represent a frequency

distribution such as that for grouped heights shown above It consists of

a set of rectangles whose areas represent the frequencies of the various

classes Often when producing these diagrams the class width is kept the

same so that the varying frequencies are represented by the height of each

rectangle When drawing histograms for grouped data the midpoints of the

rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc

An adaptation of the histogram known as the frequency polygon may also

be used to represent a frequency distribution

Example 453

Represent the data shown in Table 46 on a histogram and draw in the frequency

polygon for this distribution

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 916

Mathematics for Engineering Technicians 30

All that is required to produce the histogram is to plot frequency against the height

intervals where the intervals are drawn as class widths

Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects

the midpoint of the class widths

Another important method of representation is adding all the frequencies

of a distribution consecutively to produce a graph known as a cumulative

frequency distribution or Ogive

Figure 450 shows the cumulative frequency distribution graph for our data

given in Table 46 while Table 47 shows the consecutive addition of the

frequencies needed to produce the graph in Figure 450

From Figure 450 it is now a simple matter to find for example the median

grouped height or as it is more commonly known the 50th-percentile This

occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this

being in our case 100 giving an equivalent height of approximately 175 cm

Any percentile can be found for example the 75th-percentile where in

our case at a frequency of 150 the height can be seen to be approximately

180 cm

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

0

Class width 5 cm

Height of adults in cm

152 157 162 167 172 177 182 187 192 197

Figure 449 Figure for Example 453 histogram showing frequency distribution

KEY POINTThe frequencies of a distribution may

be added consecutively to produce a

graph known as a cumulative frequency

distribution

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians308

U N I T 4

C u m u l a t i v e F r e q u e n c y

20

30

40

50

60

70

80

90

100

110

120

130140

150

160

170

180

190

200

10

0

152 157 162 167 172 177 182

180cm

187 192 197

75th percentile

50th percentile

Figure 450 Cumulative frequency distribution graph for data given in Table 46

Table 47 Cumulative frequency data

Height (cm) Frequency Cumulative frequency

150ndash154 4 4

155ndash159 9 13

160ndash164 15 28

165ndash169 21 49

170ndash174 32 81

175ndash179 45 126

180ndash184 41 167

185ndash189 22 189

190ndash194 9 198

195ndash199 2 200

Total 200 200

TYK 411

1 In a particular university the number of students enrolled by a faculty is

given in the table below

Faculty Number of students

Business and administration 1950

Humanities and social science 2820

Physical and life sciences 1050

Technology 850

Total 6670

T e s

t yo u r

k n o

w l e

d g e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1116

Mathematics for Engineering Technicians 30

Statistical measurement

When considering statistical data it is often convenient to have one or two

values that represent the data as a whole Average values are often used

You have already found an average value when looking at the median or

50th-percentile of a cumulative frequency distribution So for example we

might talk about the average height of females in the United Kingdom being

170 cm or that the average shoe size of British males is size 9 In statistics

we may represent these average values using the mean median or mode

of the data we are considering We will spend the rest of this short section

finding these average values for both discrete and grouped data starting

with the arithmetic mean

The arithmetic mean

The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the

numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up

and divide by how many there are or more formally

Arithmetic meanarithmetic total of all the individual val

uues

number of values

n

n

sum

where the Greek symbol the sum of the individual values

x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the

data

Illustrate this data on both a bar chart and pie chart

2 For the group of numbers given below produced a tally chart and

determine their frequency of occurrence

36 41 42 38 39 40 42 41 37 40

42 44 43 41 40 38 39 39 43 39

36 37 42 38 39 42 35 42 38 39

40 41 42 37 38 39 44 45 37 40

3 Given the following frequency distribution

Class interval Frequency ( f )

60ndash64 4

65ndash69 11

70ndash74 18

75ndash79 16

80ndash84 7

85ndash90 4

(a) produce a histogram and on it draw the frequency polygon

(b) produce a cumulative frequency graph and from it determine the value

of the 50th-percentile class

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians310

U N I T 4

So for the mean of our ten numbers we have

mean

n

n

sum 8 7 9 10 5 6 12 9 6 8

10

80

108

Now no matter how long or complex the data we are dealing with provided

that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x

values rsquo is given the symbol x pronounced lsquo x bar rsquo

Example 454

The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm

163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find

the mean height of these females

Then for n 11

x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3

1118483

11168 03

x cm

Mean for grouped data

What if we are required to find the mean for grouped data Look back at

Table 46 showing the height of 200 adults grouped into ten classes In this

case the frequency of the heights needs to be taken into account

We select the class midpoint x as being the average of that class and then

multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the

frequency distribution the total value for the distribution is obtained ( fx )

This total is then divided by the sum of the frequencies ( f ) in order to

determine the mean So for grouped data

x f x f x f x f x

f f f f

f

f

n n

n

1 1 2 2 3 3

1 2 3

sdot

sdot

sumsum

( )midpoint

This rather complicated looking procedure is best illustrated by

example

Example 455

Determine the mean value for the heights of the 200 adults using the data in

Table 46

The values for each individual class are best found by producing a table using

the class midpoints and frequencies and remembering that the class midpoint is found by

dividing the sum of the upper and lower class boundaries by 2 So for example the mean

value for the 1047297rst class interval is149 5 154 5

2152

The completed table is shown

below

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 31

Midpoint ( x ) of height (cm) Frequency (f ) fx

152 4 608

157 9 1413

162 15 2430

167 21 3507

172 32 5504

177 45 7965

182 41 7462

187 22 4114

192 9 1728

197 2 394

Total f 200sum fx 35125sum

I hope you can see how each of the values was obtained Now that we have the required

totals the mean value of the distribution can be found

mean value x fx 35125

200175625 05 cm sum

sumf

Notice that our mean value of heights has the same margin of error as the original

measurements The value of the mean cannot be any more accurate than the measured

data from which it was found

Median

When some values within a set of data vary quite widely the arithmetic

mean gives a rather poor representative average of such data Under

these circumstances another more useful measure of the average is the

median

For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which

is not representative of any of the numbers given To find the median value

of the same set of numbers we simply place them in rank order that is 2

3 4 5 6 7 93 Then we select the middle (median) value Since there are

seven numbers (items) we choose the fourth item along the number 5 as

our median value

If the number of items in the set of values is even then we add together the

value of the two middle terms and divide by 2

Example 456

Find the mean and median value for the set of numbers 9 7 8 7 12 70 68

6 5 8

The arithmetic mean is found as

mean x

9 7 8 7 12 70 68 6 5 8

10

200

1020

This value is not really representative of any of the numbers in the set

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians312

U N I T 4

To 1047297nd the median value we 1047297rst put the numbers in rank order that is

5 6 7 7 8 8 912 68 70

Then from the ten numbers the two middle values The 5th and 6th values along are 8

and 8 So the medianvalue

8 8

2 8

Mode

Yet another measure of central tendency for data containing extreme

values is the mode Now the mode of a set of values containing discrete

data is the value that occurs most often So for the set of values 4 4 4 5

5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs

four times Now it is possible for a set of data to have more than one

mode For example the data used in Example 462 above has two modes

7 and 8 both of these numbers occurring twice and both occurring more

than any of the others A set of data may not have a modal value at all For

example the numbers 2 3 4 5 6 7 8 all occur once and there is

no mode

A set of data that has one mode is called unimodal data with two

modes is bimodal and data with more than two modes is known as

multimodal

When considering frequency distributions for grouped data the modal

class is that group which occurs most frequently If we wish to find

the actual modal value of a frequency distribution we need to draw a

histogram

Example 457

Find the modal class and modal value for the frequency distribution of the height

of adults given in Table 46

Referring back to Table 46 it is easy to see that the class of heights which occurs most

frequently is 175 ndash 179 cm which occurs 45 times

Now to 1047297nd the modal value we need to produce a histogram for the data

We did this for Example 453 This histogram is shown again here with the modalshown

From Figure 451 it can be seen that the modal value 17825 05 cm

This value is obtained from the intersection of the two construction lines AB and CD The

line AB is drawn diagonally from the highest value of the preceding class up to the top

right-hand corner of the modal class The line CD is drawn from the top left-hand corner

of the modal group to the lowest value of the next class immediately above the modal

group Then as can be seen the modal value is read-off where the projection line meets

the x -axis

KEY POINT

The mean median and mode are

statistical averages or measures

of central tendency for a statistical

distribution

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 31

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

C

A

B

D

0Height of adults (cm)

152 157 162 167 172 177 182 187 192 197

Modal value 17825 05 cm

Figure 451 Histogram showing frequency distribution and modal value for height of adults

TYK 412

1 Calculate the mean of the numbers 1765 986 1124 1898 959 and

888

2 Determine the mean the median and the mode for the set of numbers 9 8

7 27 16 3 1 9 4 and 116

3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the

arithmetic mean the median and the mode

4 Estimates for the length of wood required for a shelf were as follows

Length (cm) 35 36 37 38 39 40 41 42

Frequency 1 3 4 8 6 5 3 2

Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table

Length (mm) 167 168 169 170 171

Frequency 2 7 20 8 3

6 Calculate the arithmetic mean for the data shown in the table

Length of rivet (mm) 98 99 995 100 1005 101 102

Frequency 3 18 36 62 56 20 5

T e s t yo u

r k

n o

w l e

d g

e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present

Page 9: UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 30

All that is required to produce the histogram is to plot frequency against the height

intervals where the intervals are drawn as class widths

Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects

the midpoint of the class widths

Another important method of representation is adding all the frequencies

of a distribution consecutively to produce a graph known as a cumulative

frequency distribution or Ogive

Figure 450 shows the cumulative frequency distribution graph for our data

given in Table 46 while Table 47 shows the consecutive addition of the

frequencies needed to produce the graph in Figure 450

From Figure 450 it is now a simple matter to find for example the median

grouped height or as it is more commonly known the 50th-percentile This

occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this

being in our case 100 giving an equivalent height of approximately 175 cm

Any percentile can be found for example the 75th-percentile where in

our case at a frequency of 150 the height can be seen to be approximately

180 cm

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

0

Class width 5 cm

Height of adults in cm

152 157 162 167 172 177 182 187 192 197

Figure 449 Figure for Example 453 histogram showing frequency distribution

KEY POINTThe frequencies of a distribution may

be added consecutively to produce a

graph known as a cumulative frequency

distribution

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians308

U N I T 4

C u m u l a t i v e F r e q u e n c y

20

30

40

50

60

70

80

90

100

110

120

130140

150

160

170

180

190

200

10

0

152 157 162 167 172 177 182

180cm

187 192 197

75th percentile

50th percentile

Figure 450 Cumulative frequency distribution graph for data given in Table 46

Table 47 Cumulative frequency data

Height (cm) Frequency Cumulative frequency

150ndash154 4 4

155ndash159 9 13

160ndash164 15 28

165ndash169 21 49

170ndash174 32 81

175ndash179 45 126

180ndash184 41 167

185ndash189 22 189

190ndash194 9 198

195ndash199 2 200

Total 200 200

TYK 411

1 In a particular university the number of students enrolled by a faculty is

given in the table below

Faculty Number of students

Business and administration 1950

Humanities and social science 2820

Physical and life sciences 1050

Technology 850

Total 6670

T e s

t yo u r

k n o

w l e

d g e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 30

Statistical measurement

When considering statistical data it is often convenient to have one or two

values that represent the data as a whole Average values are often used

You have already found an average value when looking at the median or

50th-percentile of a cumulative frequency distribution So for example we

might talk about the average height of females in the United Kingdom being

170 cm or that the average shoe size of British males is size 9 In statistics

we may represent these average values using the mean median or mode

of the data we are considering We will spend the rest of this short section

finding these average values for both discrete and grouped data starting

with the arithmetic mean

The arithmetic mean

The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the

numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up

and divide by how many there are or more formally

Arithmetic meanarithmetic total of all the individual val

uues

number of values

n

n

sum

where the Greek symbol the sum of the individual values

x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the

data

Illustrate this data on both a bar chart and pie chart

2 For the group of numbers given below produced a tally chart and

determine their frequency of occurrence

36 41 42 38 39 40 42 41 37 40

42 44 43 41 40 38 39 39 43 39

36 37 42 38 39 42 35 42 38 39

40 41 42 37 38 39 44 45 37 40

3 Given the following frequency distribution

Class interval Frequency ( f )

60ndash64 4

65ndash69 11

70ndash74 18

75ndash79 16

80ndash84 7

85ndash90 4

(a) produce a histogram and on it draw the frequency polygon

(b) produce a cumulative frequency graph and from it determine the value

of the 50th-percentile class

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1216

Mathematics for Engineering Technicians310

U N I T 4

So for the mean of our ten numbers we have

mean

n

n

sum 8 7 9 10 5 6 12 9 6 8

10

80

108

Now no matter how long or complex the data we are dealing with provided

that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x

values rsquo is given the symbol x pronounced lsquo x bar rsquo

Example 454

The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm

163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find

the mean height of these females

Then for n 11

x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3

1118483

11168 03

x cm

Mean for grouped data

What if we are required to find the mean for grouped data Look back at

Table 46 showing the height of 200 adults grouped into ten classes In this

case the frequency of the heights needs to be taken into account

We select the class midpoint x as being the average of that class and then

multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the

frequency distribution the total value for the distribution is obtained ( fx )

This total is then divided by the sum of the frequencies ( f ) in order to

determine the mean So for grouped data

x f x f x f x f x

f f f f

f

f

n n

n

1 1 2 2 3 3

1 2 3

sdot

sdot

sumsum

( )midpoint

This rather complicated looking procedure is best illustrated by

example

Example 455

Determine the mean value for the heights of the 200 adults using the data in

Table 46

The values for each individual class are best found by producing a table using

the class midpoints and frequencies and remembering that the class midpoint is found by

dividing the sum of the upper and lower class boundaries by 2 So for example the mean

value for the 1047297rst class interval is149 5 154 5

2152

The completed table is shown

below

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 31

Midpoint ( x ) of height (cm) Frequency (f ) fx

152 4 608

157 9 1413

162 15 2430

167 21 3507

172 32 5504

177 45 7965

182 41 7462

187 22 4114

192 9 1728

197 2 394

Total f 200sum fx 35125sum

I hope you can see how each of the values was obtained Now that we have the required

totals the mean value of the distribution can be found

mean value x fx 35125

200175625 05 cm sum

sumf

Notice that our mean value of heights has the same margin of error as the original

measurements The value of the mean cannot be any more accurate than the measured

data from which it was found

Median

When some values within a set of data vary quite widely the arithmetic

mean gives a rather poor representative average of such data Under

these circumstances another more useful measure of the average is the

median

For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which

is not representative of any of the numbers given To find the median value

of the same set of numbers we simply place them in rank order that is 2

3 4 5 6 7 93 Then we select the middle (median) value Since there are

seven numbers (items) we choose the fourth item along the number 5 as

our median value

If the number of items in the set of values is even then we add together the

value of the two middle terms and divide by 2

Example 456

Find the mean and median value for the set of numbers 9 7 8 7 12 70 68

6 5 8

The arithmetic mean is found as

mean x

9 7 8 7 12 70 68 6 5 8

10

200

1020

This value is not really representative of any of the numbers in the set

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians312

U N I T 4

To 1047297nd the median value we 1047297rst put the numbers in rank order that is

5 6 7 7 8 8 912 68 70

Then from the ten numbers the two middle values The 5th and 6th values along are 8

and 8 So the medianvalue

8 8

2 8

Mode

Yet another measure of central tendency for data containing extreme

values is the mode Now the mode of a set of values containing discrete

data is the value that occurs most often So for the set of values 4 4 4 5

5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs

four times Now it is possible for a set of data to have more than one

mode For example the data used in Example 462 above has two modes

7 and 8 both of these numbers occurring twice and both occurring more

than any of the others A set of data may not have a modal value at all For

example the numbers 2 3 4 5 6 7 8 all occur once and there is

no mode

A set of data that has one mode is called unimodal data with two

modes is bimodal and data with more than two modes is known as

multimodal

When considering frequency distributions for grouped data the modal

class is that group which occurs most frequently If we wish to find

the actual modal value of a frequency distribution we need to draw a

histogram

Example 457

Find the modal class and modal value for the frequency distribution of the height

of adults given in Table 46

Referring back to Table 46 it is easy to see that the class of heights which occurs most

frequently is 175 ndash 179 cm which occurs 45 times

Now to 1047297nd the modal value we need to produce a histogram for the data

We did this for Example 453 This histogram is shown again here with the modalshown

From Figure 451 it can be seen that the modal value 17825 05 cm

This value is obtained from the intersection of the two construction lines AB and CD The

line AB is drawn diagonally from the highest value of the preceding class up to the top

right-hand corner of the modal class The line CD is drawn from the top left-hand corner

of the modal group to the lowest value of the next class immediately above the modal

group Then as can be seen the modal value is read-off where the projection line meets

the x -axis

KEY POINT

The mean median and mode are

statistical averages or measures

of central tendency for a statistical

distribution

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1516

Mathematics for Engineering Technicians 31

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

C

A

B

D

0Height of adults (cm)

152 157 162 167 172 177 182 187 192 197

Modal value 17825 05 cm

Figure 451 Histogram showing frequency distribution and modal value for height of adults

TYK 412

1 Calculate the mean of the numbers 1765 986 1124 1898 959 and

888

2 Determine the mean the median and the mode for the set of numbers 9 8

7 27 16 3 1 9 4 and 116

3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the

arithmetic mean the median and the mode

4 Estimates for the length of wood required for a shelf were as follows

Length (cm) 35 36 37 38 39 40 41 42

Frequency 1 3 4 8 6 5 3 2

Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table

Length (mm) 167 168 169 170 171

Frequency 2 7 20 8 3

6 Calculate the arithmetic mean for the data shown in the table

Length of rivet (mm) 98 99 995 100 1005 101 102

Frequency 3 18 36 62 56 20 5

T e s t yo u

r k

n o

w l e

d g

e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1616

Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present

Page 10: UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians308

U N I T 4

C u m u l a t i v e F r e q u e n c y

20

30

40

50

60

70

80

90

100

110

120

130140

150

160

170

180

190

200

10

0

152 157 162 167 172 177 182

180cm

187 192 197

75th percentile

50th percentile

Figure 450 Cumulative frequency distribution graph for data given in Table 46

Table 47 Cumulative frequency data

Height (cm) Frequency Cumulative frequency

150ndash154 4 4

155ndash159 9 13

160ndash164 15 28

165ndash169 21 49

170ndash174 32 81

175ndash179 45 126

180ndash184 41 167

185ndash189 22 189

190ndash194 9 198

195ndash199 2 200

Total 200 200

TYK 411

1 In a particular university the number of students enrolled by a faculty is

given in the table below

Faculty Number of students

Business and administration 1950

Humanities and social science 2820

Physical and life sciences 1050

Technology 850

Total 6670

T e s

t yo u r

k n o

w l e

d g e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1116

Mathematics for Engineering Technicians 30

Statistical measurement

When considering statistical data it is often convenient to have one or two

values that represent the data as a whole Average values are often used

You have already found an average value when looking at the median or

50th-percentile of a cumulative frequency distribution So for example we

might talk about the average height of females in the United Kingdom being

170 cm or that the average shoe size of British males is size 9 In statistics

we may represent these average values using the mean median or mode

of the data we are considering We will spend the rest of this short section

finding these average values for both discrete and grouped data starting

with the arithmetic mean

The arithmetic mean

The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the

numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up

and divide by how many there are or more formally

Arithmetic meanarithmetic total of all the individual val

uues

number of values

n

n

sum

where the Greek symbol the sum of the individual values

x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the

data

Illustrate this data on both a bar chart and pie chart

2 For the group of numbers given below produced a tally chart and

determine their frequency of occurrence

36 41 42 38 39 40 42 41 37 40

42 44 43 41 40 38 39 39 43 39

36 37 42 38 39 42 35 42 38 39

40 41 42 37 38 39 44 45 37 40

3 Given the following frequency distribution

Class interval Frequency ( f )

60ndash64 4

65ndash69 11

70ndash74 18

75ndash79 16

80ndash84 7

85ndash90 4

(a) produce a histogram and on it draw the frequency polygon

(b) produce a cumulative frequency graph and from it determine the value

of the 50th-percentile class

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians310

U N I T 4

So for the mean of our ten numbers we have

mean

n

n

sum 8 7 9 10 5 6 12 9 6 8

10

80

108

Now no matter how long or complex the data we are dealing with provided

that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x

values rsquo is given the symbol x pronounced lsquo x bar rsquo

Example 454

The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm

163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find

the mean height of these females

Then for n 11

x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3

1118483

11168 03

x cm

Mean for grouped data

What if we are required to find the mean for grouped data Look back at

Table 46 showing the height of 200 adults grouped into ten classes In this

case the frequency of the heights needs to be taken into account

We select the class midpoint x as being the average of that class and then

multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the

frequency distribution the total value for the distribution is obtained ( fx )

This total is then divided by the sum of the frequencies ( f ) in order to

determine the mean So for grouped data

x f x f x f x f x

f f f f

f

f

n n

n

1 1 2 2 3 3

1 2 3

sdot

sdot

sumsum

( )midpoint

This rather complicated looking procedure is best illustrated by

example

Example 455

Determine the mean value for the heights of the 200 adults using the data in

Table 46

The values for each individual class are best found by producing a table using

the class midpoints and frequencies and remembering that the class midpoint is found by

dividing the sum of the upper and lower class boundaries by 2 So for example the mean

value for the 1047297rst class interval is149 5 154 5

2152

The completed table is shown

below

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 31

Midpoint ( x ) of height (cm) Frequency (f ) fx

152 4 608

157 9 1413

162 15 2430

167 21 3507

172 32 5504

177 45 7965

182 41 7462

187 22 4114

192 9 1728

197 2 394

Total f 200sum fx 35125sum

I hope you can see how each of the values was obtained Now that we have the required

totals the mean value of the distribution can be found

mean value x fx 35125

200175625 05 cm sum

sumf

Notice that our mean value of heights has the same margin of error as the original

measurements The value of the mean cannot be any more accurate than the measured

data from which it was found

Median

When some values within a set of data vary quite widely the arithmetic

mean gives a rather poor representative average of such data Under

these circumstances another more useful measure of the average is the

median

For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which

is not representative of any of the numbers given To find the median value

of the same set of numbers we simply place them in rank order that is 2

3 4 5 6 7 93 Then we select the middle (median) value Since there are

seven numbers (items) we choose the fourth item along the number 5 as

our median value

If the number of items in the set of values is even then we add together the

value of the two middle terms and divide by 2

Example 456

Find the mean and median value for the set of numbers 9 7 8 7 12 70 68

6 5 8

The arithmetic mean is found as

mean x

9 7 8 7 12 70 68 6 5 8

10

200

1020

This value is not really representative of any of the numbers in the set

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians312

U N I T 4

To 1047297nd the median value we 1047297rst put the numbers in rank order that is

5 6 7 7 8 8 912 68 70

Then from the ten numbers the two middle values The 5th and 6th values along are 8

and 8 So the medianvalue

8 8

2 8

Mode

Yet another measure of central tendency for data containing extreme

values is the mode Now the mode of a set of values containing discrete

data is the value that occurs most often So for the set of values 4 4 4 5

5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs

four times Now it is possible for a set of data to have more than one

mode For example the data used in Example 462 above has two modes

7 and 8 both of these numbers occurring twice and both occurring more

than any of the others A set of data may not have a modal value at all For

example the numbers 2 3 4 5 6 7 8 all occur once and there is

no mode

A set of data that has one mode is called unimodal data with two

modes is bimodal and data with more than two modes is known as

multimodal

When considering frequency distributions for grouped data the modal

class is that group which occurs most frequently If we wish to find

the actual modal value of a frequency distribution we need to draw a

histogram

Example 457

Find the modal class and modal value for the frequency distribution of the height

of adults given in Table 46

Referring back to Table 46 it is easy to see that the class of heights which occurs most

frequently is 175 ndash 179 cm which occurs 45 times

Now to 1047297nd the modal value we need to produce a histogram for the data

We did this for Example 453 This histogram is shown again here with the modalshown

From Figure 451 it can be seen that the modal value 17825 05 cm

This value is obtained from the intersection of the two construction lines AB and CD The

line AB is drawn diagonally from the highest value of the preceding class up to the top

right-hand corner of the modal class The line CD is drawn from the top left-hand corner

of the modal group to the lowest value of the next class immediately above the modal

group Then as can be seen the modal value is read-off where the projection line meets

the x -axis

KEY POINT

The mean median and mode are

statistical averages or measures

of central tendency for a statistical

distribution

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 31

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

C

A

B

D

0Height of adults (cm)

152 157 162 167 172 177 182 187 192 197

Modal value 17825 05 cm

Figure 451 Histogram showing frequency distribution and modal value for height of adults

TYK 412

1 Calculate the mean of the numbers 1765 986 1124 1898 959 and

888

2 Determine the mean the median and the mode for the set of numbers 9 8

7 27 16 3 1 9 4 and 116

3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the

arithmetic mean the median and the mode

4 Estimates for the length of wood required for a shelf were as follows

Length (cm) 35 36 37 38 39 40 41 42

Frequency 1 3 4 8 6 5 3 2

Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table

Length (mm) 167 168 169 170 171

Frequency 2 7 20 8 3

6 Calculate the arithmetic mean for the data shown in the table

Length of rivet (mm) 98 99 995 100 1005 101 102

Frequency 3 18 36 62 56 20 5

T e s t yo u

r k

n o

w l e

d g

e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present

Page 11: UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 30

Statistical measurement

When considering statistical data it is often convenient to have one or two

values that represent the data as a whole Average values are often used

You have already found an average value when looking at the median or

50th-percentile of a cumulative frequency distribution So for example we

might talk about the average height of females in the United Kingdom being

170 cm or that the average shoe size of British males is size 9 In statistics

we may represent these average values using the mean median or mode

of the data we are considering We will spend the rest of this short section

finding these average values for both discrete and grouped data starting

with the arithmetic mean

The arithmetic mean

The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the

numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up

and divide by how many there are or more formally

Arithmetic meanarithmetic total of all the individual val

uues

number of values

n

n

sum

where the Greek symbol the sum of the individual values

x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the

data

Illustrate this data on both a bar chart and pie chart

2 For the group of numbers given below produced a tally chart and

determine their frequency of occurrence

36 41 42 38 39 40 42 41 37 40

42 44 43 41 40 38 39 39 43 39

36 37 42 38 39 42 35 42 38 39

40 41 42 37 38 39 44 45 37 40

3 Given the following frequency distribution

Class interval Frequency ( f )

60ndash64 4

65ndash69 11

70ndash74 18

75ndash79 16

80ndash84 7

85ndash90 4

(a) produce a histogram and on it draw the frequency polygon

(b) produce a cumulative frequency graph and from it determine the value

of the 50th-percentile class

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians310

U N I T 4

So for the mean of our ten numbers we have

mean

n

n

sum 8 7 9 10 5 6 12 9 6 8

10

80

108

Now no matter how long or complex the data we are dealing with provided

that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x

values rsquo is given the symbol x pronounced lsquo x bar rsquo

Example 454

The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm

163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find

the mean height of these females

Then for n 11

x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3

1118483

11168 03

x cm

Mean for grouped data

What if we are required to find the mean for grouped data Look back at

Table 46 showing the height of 200 adults grouped into ten classes In this

case the frequency of the heights needs to be taken into account

We select the class midpoint x as being the average of that class and then

multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the

frequency distribution the total value for the distribution is obtained ( fx )

This total is then divided by the sum of the frequencies ( f ) in order to

determine the mean So for grouped data

x f x f x f x f x

f f f f

f

f

n n

n

1 1 2 2 3 3

1 2 3

sdot

sdot

sumsum

( )midpoint

This rather complicated looking procedure is best illustrated by

example

Example 455

Determine the mean value for the heights of the 200 adults using the data in

Table 46

The values for each individual class are best found by producing a table using

the class midpoints and frequencies and remembering that the class midpoint is found by

dividing the sum of the upper and lower class boundaries by 2 So for example the mean

value for the 1047297rst class interval is149 5 154 5

2152

The completed table is shown

below

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 31

Midpoint ( x ) of height (cm) Frequency (f ) fx

152 4 608

157 9 1413

162 15 2430

167 21 3507

172 32 5504

177 45 7965

182 41 7462

187 22 4114

192 9 1728

197 2 394

Total f 200sum fx 35125sum

I hope you can see how each of the values was obtained Now that we have the required

totals the mean value of the distribution can be found

mean value x fx 35125

200175625 05 cm sum

sumf

Notice that our mean value of heights has the same margin of error as the original

measurements The value of the mean cannot be any more accurate than the measured

data from which it was found

Median

When some values within a set of data vary quite widely the arithmetic

mean gives a rather poor representative average of such data Under

these circumstances another more useful measure of the average is the

median

For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which

is not representative of any of the numbers given To find the median value

of the same set of numbers we simply place them in rank order that is 2

3 4 5 6 7 93 Then we select the middle (median) value Since there are

seven numbers (items) we choose the fourth item along the number 5 as

our median value

If the number of items in the set of values is even then we add together the

value of the two middle terms and divide by 2

Example 456

Find the mean and median value for the set of numbers 9 7 8 7 12 70 68

6 5 8

The arithmetic mean is found as

mean x

9 7 8 7 12 70 68 6 5 8

10

200

1020

This value is not really representative of any of the numbers in the set

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians312

U N I T 4

To 1047297nd the median value we 1047297rst put the numbers in rank order that is

5 6 7 7 8 8 912 68 70

Then from the ten numbers the two middle values The 5th and 6th values along are 8

and 8 So the medianvalue

8 8

2 8

Mode

Yet another measure of central tendency for data containing extreme

values is the mode Now the mode of a set of values containing discrete

data is the value that occurs most often So for the set of values 4 4 4 5

5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs

four times Now it is possible for a set of data to have more than one

mode For example the data used in Example 462 above has two modes

7 and 8 both of these numbers occurring twice and both occurring more

than any of the others A set of data may not have a modal value at all For

example the numbers 2 3 4 5 6 7 8 all occur once and there is

no mode

A set of data that has one mode is called unimodal data with two

modes is bimodal and data with more than two modes is known as

multimodal

When considering frequency distributions for grouped data the modal

class is that group which occurs most frequently If we wish to find

the actual modal value of a frequency distribution we need to draw a

histogram

Example 457

Find the modal class and modal value for the frequency distribution of the height

of adults given in Table 46

Referring back to Table 46 it is easy to see that the class of heights which occurs most

frequently is 175 ndash 179 cm which occurs 45 times

Now to 1047297nd the modal value we need to produce a histogram for the data

We did this for Example 453 This histogram is shown again here with the modalshown

From Figure 451 it can be seen that the modal value 17825 05 cm

This value is obtained from the intersection of the two construction lines AB and CD The

line AB is drawn diagonally from the highest value of the preceding class up to the top

right-hand corner of the modal class The line CD is drawn from the top left-hand corner

of the modal group to the lowest value of the next class immediately above the modal

group Then as can be seen the modal value is read-off where the projection line meets

the x -axis

KEY POINT

The mean median and mode are

statistical averages or measures

of central tendency for a statistical

distribution

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 31

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

C

A

B

D

0Height of adults (cm)

152 157 162 167 172 177 182 187 192 197

Modal value 17825 05 cm

Figure 451 Histogram showing frequency distribution and modal value for height of adults

TYK 412

1 Calculate the mean of the numbers 1765 986 1124 1898 959 and

888

2 Determine the mean the median and the mode for the set of numbers 9 8

7 27 16 3 1 9 4 and 116

3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the

arithmetic mean the median and the mode

4 Estimates for the length of wood required for a shelf were as follows

Length (cm) 35 36 37 38 39 40 41 42

Frequency 1 3 4 8 6 5 3 2

Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table

Length (mm) 167 168 169 170 171

Frequency 2 7 20 8 3

6 Calculate the arithmetic mean for the data shown in the table

Length of rivet (mm) 98 99 995 100 1005 101 102

Frequency 3 18 36 62 56 20 5

T e s t yo u

r k

n o

w l e

d g

e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present

Page 12: UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1216

Mathematics for Engineering Technicians310

U N I T 4

So for the mean of our ten numbers we have

mean

n

n

sum 8 7 9 10 5 6 12 9 6 8

10

80

108

Now no matter how long or complex the data we are dealing with provided

that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x

values rsquo is given the symbol x pronounced lsquo x bar rsquo

Example 454

The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm

163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find

the mean height of these females

Then for n 11

x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3

1118483

11168 03

x cm

Mean for grouped data

What if we are required to find the mean for grouped data Look back at

Table 46 showing the height of 200 adults grouped into ten classes In this

case the frequency of the heights needs to be taken into account

We select the class midpoint x as being the average of that class and then

multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the

frequency distribution the total value for the distribution is obtained ( fx )

This total is then divided by the sum of the frequencies ( f ) in order to

determine the mean So for grouped data

x f x f x f x f x

f f f f

f

f

n n

n

1 1 2 2 3 3

1 2 3

sdot

sdot

sumsum

( )midpoint

This rather complicated looking procedure is best illustrated by

example

Example 455

Determine the mean value for the heights of the 200 adults using the data in

Table 46

The values for each individual class are best found by producing a table using

the class midpoints and frequencies and remembering that the class midpoint is found by

dividing the sum of the upper and lower class boundaries by 2 So for example the mean

value for the 1047297rst class interval is149 5 154 5

2152

The completed table is shown

below

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 31

Midpoint ( x ) of height (cm) Frequency (f ) fx

152 4 608

157 9 1413

162 15 2430

167 21 3507

172 32 5504

177 45 7965

182 41 7462

187 22 4114

192 9 1728

197 2 394

Total f 200sum fx 35125sum

I hope you can see how each of the values was obtained Now that we have the required

totals the mean value of the distribution can be found

mean value x fx 35125

200175625 05 cm sum

sumf

Notice that our mean value of heights has the same margin of error as the original

measurements The value of the mean cannot be any more accurate than the measured

data from which it was found

Median

When some values within a set of data vary quite widely the arithmetic

mean gives a rather poor representative average of such data Under

these circumstances another more useful measure of the average is the

median

For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which

is not representative of any of the numbers given To find the median value

of the same set of numbers we simply place them in rank order that is 2

3 4 5 6 7 93 Then we select the middle (median) value Since there are

seven numbers (items) we choose the fourth item along the number 5 as

our median value

If the number of items in the set of values is even then we add together the

value of the two middle terms and divide by 2

Example 456

Find the mean and median value for the set of numbers 9 7 8 7 12 70 68

6 5 8

The arithmetic mean is found as

mean x

9 7 8 7 12 70 68 6 5 8

10

200

1020

This value is not really representative of any of the numbers in the set

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians312

U N I T 4

To 1047297nd the median value we 1047297rst put the numbers in rank order that is

5 6 7 7 8 8 912 68 70

Then from the ten numbers the two middle values The 5th and 6th values along are 8

and 8 So the medianvalue

8 8

2 8

Mode

Yet another measure of central tendency for data containing extreme

values is the mode Now the mode of a set of values containing discrete

data is the value that occurs most often So for the set of values 4 4 4 5

5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs

four times Now it is possible for a set of data to have more than one

mode For example the data used in Example 462 above has two modes

7 and 8 both of these numbers occurring twice and both occurring more

than any of the others A set of data may not have a modal value at all For

example the numbers 2 3 4 5 6 7 8 all occur once and there is

no mode

A set of data that has one mode is called unimodal data with two

modes is bimodal and data with more than two modes is known as

multimodal

When considering frequency distributions for grouped data the modal

class is that group which occurs most frequently If we wish to find

the actual modal value of a frequency distribution we need to draw a

histogram

Example 457

Find the modal class and modal value for the frequency distribution of the height

of adults given in Table 46

Referring back to Table 46 it is easy to see that the class of heights which occurs most

frequently is 175 ndash 179 cm which occurs 45 times

Now to 1047297nd the modal value we need to produce a histogram for the data

We did this for Example 453 This histogram is shown again here with the modalshown

From Figure 451 it can be seen that the modal value 17825 05 cm

This value is obtained from the intersection of the two construction lines AB and CD The

line AB is drawn diagonally from the highest value of the preceding class up to the top

right-hand corner of the modal class The line CD is drawn from the top left-hand corner

of the modal group to the lowest value of the next class immediately above the modal

group Then as can be seen the modal value is read-off where the projection line meets

the x -axis

KEY POINT

The mean median and mode are

statistical averages or measures

of central tendency for a statistical

distribution

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 31

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

C

A

B

D

0Height of adults (cm)

152 157 162 167 172 177 182 187 192 197

Modal value 17825 05 cm

Figure 451 Histogram showing frequency distribution and modal value for height of adults

TYK 412

1 Calculate the mean of the numbers 1765 986 1124 1898 959 and

888

2 Determine the mean the median and the mode for the set of numbers 9 8

7 27 16 3 1 9 4 and 116

3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the

arithmetic mean the median and the mode

4 Estimates for the length of wood required for a shelf were as follows

Length (cm) 35 36 37 38 39 40 41 42

Frequency 1 3 4 8 6 5 3 2

Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table

Length (mm) 167 168 169 170 171

Frequency 2 7 20 8 3

6 Calculate the arithmetic mean for the data shown in the table

Length of rivet (mm) 98 99 995 100 1005 101 102

Frequency 3 18 36 62 56 20 5

T e s t yo u

r k

n o

w l e

d g

e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present

Page 13: UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 31

Midpoint ( x ) of height (cm) Frequency (f ) fx

152 4 608

157 9 1413

162 15 2430

167 21 3507

172 32 5504

177 45 7965

182 41 7462

187 22 4114

192 9 1728

197 2 394

Total f 200sum fx 35125sum

I hope you can see how each of the values was obtained Now that we have the required

totals the mean value of the distribution can be found

mean value x fx 35125

200175625 05 cm sum

sumf

Notice that our mean value of heights has the same margin of error as the original

measurements The value of the mean cannot be any more accurate than the measured

data from which it was found

Median

When some values within a set of data vary quite widely the arithmetic

mean gives a rather poor representative average of such data Under

these circumstances another more useful measure of the average is the

median

For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which

is not representative of any of the numbers given To find the median value

of the same set of numbers we simply place them in rank order that is 2

3 4 5 6 7 93 Then we select the middle (median) value Since there are

seven numbers (items) we choose the fourth item along the number 5 as

our median value

If the number of items in the set of values is even then we add together the

value of the two middle terms and divide by 2

Example 456

Find the mean and median value for the set of numbers 9 7 8 7 12 70 68

6 5 8

The arithmetic mean is found as

mean x

9 7 8 7 12 70 68 6 5 8

10

200

1020

This value is not really representative of any of the numbers in the set

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians312

U N I T 4

To 1047297nd the median value we 1047297rst put the numbers in rank order that is

5 6 7 7 8 8 912 68 70

Then from the ten numbers the two middle values The 5th and 6th values along are 8

and 8 So the medianvalue

8 8

2 8

Mode

Yet another measure of central tendency for data containing extreme

values is the mode Now the mode of a set of values containing discrete

data is the value that occurs most often So for the set of values 4 4 4 5

5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs

four times Now it is possible for a set of data to have more than one

mode For example the data used in Example 462 above has two modes

7 and 8 both of these numbers occurring twice and both occurring more

than any of the others A set of data may not have a modal value at all For

example the numbers 2 3 4 5 6 7 8 all occur once and there is

no mode

A set of data that has one mode is called unimodal data with two

modes is bimodal and data with more than two modes is known as

multimodal

When considering frequency distributions for grouped data the modal

class is that group which occurs most frequently If we wish to find

the actual modal value of a frequency distribution we need to draw a

histogram

Example 457

Find the modal class and modal value for the frequency distribution of the height

of adults given in Table 46

Referring back to Table 46 it is easy to see that the class of heights which occurs most

frequently is 175 ndash 179 cm which occurs 45 times

Now to 1047297nd the modal value we need to produce a histogram for the data

We did this for Example 453 This histogram is shown again here with the modalshown

From Figure 451 it can be seen that the modal value 17825 05 cm

This value is obtained from the intersection of the two construction lines AB and CD The

line AB is drawn diagonally from the highest value of the preceding class up to the top

right-hand corner of the modal class The line CD is drawn from the top left-hand corner

of the modal group to the lowest value of the next class immediately above the modal

group Then as can be seen the modal value is read-off where the projection line meets

the x -axis

KEY POINT

The mean median and mode are

statistical averages or measures

of central tendency for a statistical

distribution

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 31

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

C

A

B

D

0Height of adults (cm)

152 157 162 167 172 177 182 187 192 197

Modal value 17825 05 cm

Figure 451 Histogram showing frequency distribution and modal value for height of adults

TYK 412

1 Calculate the mean of the numbers 1765 986 1124 1898 959 and

888

2 Determine the mean the median and the mode for the set of numbers 9 8

7 27 16 3 1 9 4 and 116

3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the

arithmetic mean the median and the mode

4 Estimates for the length of wood required for a shelf were as follows

Length (cm) 35 36 37 38 39 40 41 42

Frequency 1 3 4 8 6 5 3 2

Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table

Length (mm) 167 168 169 170 171

Frequency 2 7 20 8 3

6 Calculate the arithmetic mean for the data shown in the table

Length of rivet (mm) 98 99 995 100 1005 101 102

Frequency 3 18 36 62 56 20 5

T e s t yo u

r k

n o

w l e

d g

e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1616

Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present

Page 14: UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians312

U N I T 4

To 1047297nd the median value we 1047297rst put the numbers in rank order that is

5 6 7 7 8 8 912 68 70

Then from the ten numbers the two middle values The 5th and 6th values along are 8

and 8 So the medianvalue

8 8

2 8

Mode

Yet another measure of central tendency for data containing extreme

values is the mode Now the mode of a set of values containing discrete

data is the value that occurs most often So for the set of values 4 4 4 5

5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs

four times Now it is possible for a set of data to have more than one

mode For example the data used in Example 462 above has two modes

7 and 8 both of these numbers occurring twice and both occurring more

than any of the others A set of data may not have a modal value at all For

example the numbers 2 3 4 5 6 7 8 all occur once and there is

no mode

A set of data that has one mode is called unimodal data with two

modes is bimodal and data with more than two modes is known as

multimodal

When considering frequency distributions for grouped data the modal

class is that group which occurs most frequently If we wish to find

the actual modal value of a frequency distribution we need to draw a

histogram

Example 457

Find the modal class and modal value for the frequency distribution of the height

of adults given in Table 46

Referring back to Table 46 it is easy to see that the class of heights which occurs most

frequently is 175 ndash 179 cm which occurs 45 times

Now to 1047297nd the modal value we need to produce a histogram for the data

We did this for Example 453 This histogram is shown again here with the modalshown

From Figure 451 it can be seen that the modal value 17825 05 cm

This value is obtained from the intersection of the two construction lines AB and CD The

line AB is drawn diagonally from the highest value of the preceding class up to the top

right-hand corner of the modal class The line CD is drawn from the top left-hand corner

of the modal group to the lowest value of the next class immediately above the modal

group Then as can be seen the modal value is read-off where the projection line meets

the x -axis

KEY POINT

The mean median and mode are

statistical averages or measures

of central tendency for a statistical

distribution

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 31

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

C

A

B

D

0Height of adults (cm)

152 157 162 167 172 177 182 187 192 197

Modal value 17825 05 cm

Figure 451 Histogram showing frequency distribution and modal value for height of adults

TYK 412

1 Calculate the mean of the numbers 1765 986 1124 1898 959 and

888

2 Determine the mean the median and the mode for the set of numbers 9 8

7 27 16 3 1 9 4 and 116

3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the

arithmetic mean the median and the mode

4 Estimates for the length of wood required for a shelf were as follows

Length (cm) 35 36 37 38 39 40 41 42

Frequency 1 3 4 8 6 5 3 2

Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table

Length (mm) 167 168 169 170 171

Frequency 2 7 20 8 3

6 Calculate the arithmetic mean for the data shown in the table

Length of rivet (mm) 98 99 995 100 1005 101 102

Frequency 3 18 36 62 56 20 5

T e s t yo u

r k

n o

w l e

d g

e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1616

Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present

Page 15: UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

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Mathematics for Engineering Technicians 31

F r e q u e n c y

50

45

40

35

30

25

20

15

10

5

C

A

B

D

0Height of adults (cm)

152 157 162 167 172 177 182 187 192 197

Modal value 17825 05 cm

Figure 451 Histogram showing frequency distribution and modal value for height of adults

TYK 412

1 Calculate the mean of the numbers 1765 986 1124 1898 959 and

888

2 Determine the mean the median and the mode for the set of numbers 9 8

7 27 16 3 1 9 4 and 116

3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the

arithmetic mean the median and the mode

4 Estimates for the length of wood required for a shelf were as follows

Length (cm) 35 36 37 38 39 40 41 42

Frequency 1 3 4 8 6 5 3 2

Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table

Length (mm) 167 168 169 170 171

Frequency 2 7 20 8 3

6 Calculate the arithmetic mean for the data shown in the table

Length of rivet (mm) 98 99 995 100 1005 101 102

Frequency 3 18 36 62 56 20 5

T e s t yo u

r k

n o

w l e

d g

e

TYK

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1616

Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present

Page 16: UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3

httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1616

Mathematics for Engineering Technicians314

U N I T 4

Elementary Calculus Techniques

Introduction

Meeting the calculus for the first time is often a rather daunting business

In order to appreciate the power of this branch of mathematics we

must first attempt to define it So what is the calculus and what is its

function

Imagine driving a car or riding a motorcycle starting from rest over a

measured distance say 1 km If your time for the run was 25 seconds then

we can find your average speed over the measured kilometre from the

fact that speed distancetime Then using consistent units your average

speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were

testing the vehicle and we needed to know its acceleration after you had

driven 500 m In order to find this we would need to determine how the

vehicle speed was changing at this exact point because the rate at which

your vehicle speed changes is its acceleration To find things such as rate of

change of speed we can use calculus techniques

The calculus is split into two major areas the differential calculus and the

integral calculus

The differential calculus is a branch of mathematics concerned with finding

how things change with respect to variables such as time distance or speed

especially when these changes are continually varying In engineering we

are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure

density and temperature change with height or time Also how electrical

quantities vary with time such as electrical charge alternating current

electrical power etc All these areas may be investigated using the

differential calculus

The integral calculus has two primary functions It can be used to find

the length of arcs surface areas or volumes enclosed by a surface Its

second function is that of anti-differentiation For example we can use the

differential calculus to find the rate of change of distance of our motorcycle

7 Tests were carried out on 50 occasions to determine the percentage of

greenhouse gases in the emissions from an internal combustion engine

The results from the tests showing the percentage of greenhouse gases

recorded were as follows

greenhouse gases present 32 33 34 35 36 37

Frequency 2 12 20 8 6 2

(a) Determine the arithmetic mean for the greenhouse gases present

(b) Produce a histogram for the data and from it find an estimate for the

modal value

(c) Produce a cumulative frequency distribution curve and from it determine

the median value of greenhouse gases present