ba201 engineering mathematic unit1 - statistics

7/21/2019 BA201 Engineering Mathematic UNIT1 - Statistics

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Prepared by : NURIZATY BINTI MUHAMAD NOR Page 1

Statistics

1.1Explain the basic terminologies of statistics

1.2Explain several forms of data presentation

1.3Determine class interval, upper limit, lower

limit, class size, middle value and others.1.4Prepare a frequency table

By the end of this chapter, you should be able to

1. Construct pictograms, bar charts, linegraphs and pie-chart to represent

data.

2. Organise data by constructing

frequency tables

3. Construct a histogram, frequency

polygon and ogive.

General Ob ectives

S ecific Ob ectives


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1.0INTRODUCTION

Statistics can be defined as the science of collecting, organizing, presenting, analyzing and

interpreting data to assist in making more effective decisions.

1.1 REPRESENTING DATA

Charts, pictures and graphs are representations of some data with is easy to look at. In common

usage, one can gain useful information by using various graphs.

1.1.1 PICTOGRAM

A pictogram is a representation of some data using symbols to show the frequency of something

There are advantages and disadvantages in using pictograms to represent data.

a) Advantages :

i) more attractive

ii) easy to understand

iii) easy to remember

b) Disadvantages :

i) Problem in drawing similar pictures or symbols

ii) Value that is represented by a particular symbol has to be memorized

iii) Represented value may not be accurate

Constructing pictograms

Step 1 : Analyse the information

Step 2 : Decide on a suitable symbol to represent the data

Step 3 : Decide on the key or symbol

Step 4 : Prepare the table

Step 5 : Write the title.

INPUT


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Example 1.1

Construct a pictogram according to this following data in Table 1.1

Table 1.1 : Development of Housing Sector in Malaysia

Years Unit1981 686101982 87810

1983 72760

1984 830801985 93810

Solutions :

Pictogram : Development of Housing Sector in Malaysia

1981 1982 1983 1984 1985

Key : represents 10000 units

Example 1.2 :

The pictogram shows the fines collected for the late return of library books.

Collection of fines

Monday

TuesdayWednesday

Thursday

Key : represents RM5.00


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a) What is the total sum collected as fines?

= (11+1.5+1.5) x RM5.00 = RM60.00 (total number of symbols x value of 1 symbol)

b) If RM22.50 was collected on Friday, how many symbols must be drawn for this sum?

= sum of money collected / value of 1 symbol

= RM22.50 / RM5.00

= 4.5 symbols.

1.1.2 LINE GRAPH

A line graph represents data that is obtained over a period of time by drawing straight lines

which join the coordinates given by data. The horizontal axis represents the period of time.

There are advantages and disadvantages in using a line graph to represent data.

a) Advantages:

i) Able to trace the change in data over a specific period of time

ii) The value of the data can be shown more accurately

iii) Can be used to estimate

b) Disadvantages :

i) Not easy to interpret

ii) Not attractive

Constructing a line graph

Step 1 : Decide on an appropriate scale for both axes

Step 2 : Draw the two axes and plot the coordinates

Step 3 : Join all the points with a ruler

Step 4 : Give the suitable title for the line graph


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Example 1.2

Draw a line graph to represent the given data in Table 1.2.

Table 1.2 Saving balance in Maybank Berhad

Year Balance (million)

1980 241981 321982 28

1983 37

1984 441985 53

1986 50

Solution:

Figure 1.1: A Line Graph of Saving balance in Maybank Berhad , 1980 - 1986

1.1.3 BAR CHART

A bar chartis a representation of data using either verticalor horizontal bars. It isactually a

frequency diagram using rectangles of equal width with height or length are proportional to the

frequency.

There are advantages and disadvantages in using a bar graph to represent data.

Year

Balance

(million)


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a) Advantages:

i) Easy to draw

ii) Data value is more accurately shown

iii) Information can be obtained immediately from the chart

iv) Easy to compare values

b) Disadvantages :

i) Each component cannot be compared with the whole

Example 1.3:

Construct a vertical bar chart and horizontal bar based on data in Table 1.3 given.

Table 1.3: Total Import for West Malaysia

Year Sum in million (RM)

1971 3.405

1972 3.8771973 5.143

1974 8.5501975 7.496

Constructing a bar chart

Step 1 : Decide on an appropriate scale for both axes

Step 2 : Draw and label the vertical and horizontal axes

Step 3 : Ascertain the length of the bar

Step 4 : Draw bars of equal length vertically/horizontally



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Solution :

:

Figure 1.3 : Horizontal Bar Chart

0 2 4 6 8 10

1971

1972

1973

1974

1975

Year

RM 1000 million

Bar Chart : Total Import for West Malaysia


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1.1.4 PIE-CHART

A pie-chart represents relative quantities by size of sectors of a circle.

Percentage of sector = frequency of data x 100%

Total frequency

Angle of sector = frequency of data x 360

Total frequency

Example 1.4

Draw a pie-chart based on data in table 1.4.

Jadual 1.4 : Grade scored by 128 students in a statistic course

Grade Number of

Students

A 12

B 33

C 34

D 33E 16

Constructing pie-chart

Step 1 : Construct a table by analyse the information in percentage and angle of sector

Step 2 : Write each sector representing data. Use a key if sector in the pie chart is small.

Step 3 : Arrange the sector of data in a clockwise



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Solution:

Step 1 : Construct a table

Grade Number of

Students

Percentage (%) Angle (o)

A 12 9.38 33.75

B 33 25.78 92.81

C 34 26.56 95.62

D 33 25.78 92.81

E 16 12.50 45.00

Step 2 : Draw a pie-chart

26.56%

25.78%

9.38%12.5%

25.78%

GradeA

Grade

B

GradeE

Grade

D

Grade

C


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ACTIVITY 1a

1a.1 Draw a vertical bar graph using the following data which shows the age of 80 workers in

Company JJ Sdn. Bhd.

Age (year) Number of workers

20-24 8

25-29 12

30-34 14

35-39 1740-44 13

45-49 950-54 4

55-59 3

1a.2 Construct a pie-chart using the following data of Petroleum Reserve by Countries in WesternHemisphere.

Country Percentage

Canada 6.8Mexico 46.1

South America 29.7

USA 17.4

1a.3 Construct a line graph using the data below.

Week Sales of motorcycle

1 6

2 10

3 9

4 11

5 15

6 13

7 16

8 12


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ANSWERS 1a

1a.1

1a.2

1a.3

6.8%

46.1%29.7%

17.4%

0

5

10

15

20

1 2 3 4 5 6 7 8salesofmotorcycle

Weeks

Number of workers

Ages ( Year )


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1.2FREQUENCY DISTRIBUTION TABLE

A set of a raw data which consists of many measurements of a certain quantity can be grouped

into several classes. The range of values of each class is known as the class interval.Now, let us discuss

how to make a frequency distribution table for the grouped data.

For example, the weight of each of the 40 students in the mathematics class is shown as below :

Table 1.5: The weight of 40 students in the mathematics class45 50 55 46 46 51 54 60 62 64

58 48 51 56 48 47 50 53 53 60

59 49 61 48 59 60 50 53 52 53

55 56 49 50 61 63 49 54 54 56

How to construct a frequency distribution table using a raw datagiven.

1. Determine the range. (highest value of data - lowest value of data)

The highest value = 64kg

The lowest value = 45kg

So, the range is (64 -45) = 19 kg.

2. Determine thenumber of class interval.

The number of class interval usually between 5 and 20, in such a way that the lowest value is

included in its first class interval and the highest value must be included in the last class interval.

One rule that can help to decide on the number of clases using Sturges Formula :

C = 1 + 3.3 log N

Where c = number of classes

N = The total number of observations in the data set.

INPUT


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Example : c = 1 + 3.3 log 40

= 6.29

3. Determine the width of class interval (size of class interval)

Size of class interval = range / number of class interval

Example : range = 19 kg

Number of class interval = 6

So, size of class interval = 19/6 = 3.17 ( convert to the nearest of highest number)

= 4kg

So, the class interval : (45- 48) , (49 - 52) , (53 - 56) , (57 - 60) , ( 61 - 64)

4. Determine the boundary of a class interval

Lower boundary is the midpoint between the lower limit of class interval and the upper limit of

previous class interval.

Upper boundary is the midpoint between the upper limit of class interval and the lower limit of

succeeding class interval.

previous class interval succeeding class interval

(45- 48) (49 - 52) (53 - 56)

Lower boundary Upper boundary

= (48+49) = (52 + 53)

= 48.5 = 52.5

Class boundary

44.5 - 48.5

48.5 - 52.5

52.5 - 56.5

56.5 - 60.5

60.5 - 64.5

Table 1.6


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Table 1.7 : Frequency distribution of 40 students of mathematics class

Weight (kg) Tally Frequency

44.5 - 48.5

48.5 - 52.5

52.5 - 56.5

56.5 - 60.5

60.5 - 64.5

IIII II

IIII IIII

IIII IIII II

IIII I

IIII

7

10

12

6

5

1.3HISTOGRAM, POLYGON FREQUENCY AND OGIF

1.3.1 Histogram

A histogram is a graphical representation of a frequency table. A histogram is a vertical bar chart

without any spacing between the bars.

How to construct a histogram :

Find the lower and upper boundaries of each class interval for

example from table 1.7 : 44.5 - 48.5, 48.5 - 52.5,..

Choose suitable scales to represent the size of class interval on the

horizontal axis and the frequencies on the vertical axis.

Draw a rectangle for each class interval, with its width usually

representing the size and its height representing the frequency of the

class interval.

Write the title of the histogram.


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Example 1.5

Draw a histogram using the data in table 1.7.

Solution :

Rajah 1.5 : A histogram of the weight of 40 students.

1.3.2 Polygon Frequency

A Polygon Frequency is the line chart of a frequency distribution.

7

10

12

6

5

0

2

4

6

8

10

12

14

F

requensy

44.5 48.5 52.5 56.5 60.5 64.5

Masses (kg)

Frequency

How to construct a polygon frequency based on the above histogram.:

1. Determine the midpoints. Example from table 1.5, the first class intervals :

44.5 - 48.5, the midpoint : (44.5 + 48.5)/2 = 46.5 .Mark the midpoints at the tops of

the rectangular bars in the histogram.

2. Add two class intervals with zero frequency to the histogram, one before the first

class interval and one after the last class interval. Mark the midpoints of these two

class intervals.

3. Draw straight lines joining the midpoints of the consecutive rectangular bars.

4. Write the title of the graph.


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Example 1.6

Draw a polygon frequency from table 1.7.

Solution :

1.3.3 Ogive

An ogive or cumulative frequency graph is a graphical representation of a cumulative

frequency distribution. A cumulative frequency graph is produced by plotting the cumulative

frequency of each class interval against its upper boundary and then joining the points with a

smooth curve.

There are two types of ogive that is

1) more than ogive

2) less than ogive

0

2

4

6

8

10

12

14

Frequency

Figure 1.6 : Polygon Frequency For The Weight of 40students

44.5 48.5 52.5 56.5 60.5 64.5

Weight (kg)

How to construct a polygon frequency based on the above histogram.:

1. Construct a table which shows the cumulative frequencies and the upper boundaries

of the data.

2. Select suitable scales to represent cumulative frequency on the vertical axis and

upper boundary on the horizontal axis.

3. Plot the values of cumulative frequency against the values of upper boundary. Then

draw a smooth curve through the successive points plotted.

4. Write the title of the histogram.


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Example 1.7

Construct a more thanogive and a less than ogiveusing data in table 1.7.

Solution :

Table 1.8: Frequency distribution less than

Weight Cumulative Frequency

Less than 44.5

Less than 48.5

Less than 52.5

Less than 56.5

Less than 60.5

Less than 64.5

0

7

18 (7 + 11)

29 (7 + 11 + 11)

35 (7 + 11 + 11 + 6)

40 (7 + 11 + 11 + 6 + 5)

Table 1.9 : Frequency distribution more than

Weight Cumulative Frequency

More than 64.5

More than 60.5

More than 56.5More than 52.5

More than 48.5

More than 44.5

40 (7 + 11 + 11 + 6 + 5)

35 (7 + 11 + 11 + 6)

29 (7 + 11 + 11)18 (7 + 11)

7

0


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Figure 1.7 : A less than andmore than ogive of the weightof 40 students.


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ACTIVITY 1b

1b.1 Construct the ogive less than and more than using the following data in table 1.10. From theogive, estimate the number of students that are less than 125 pound and the number of students

that are more than 140 pounds.

Table 1.10

Class Frequency

95.5113.5 5

113.5131.5 7

131.5149.5 5

149.5167.5 2

167.5185.5 5

24

1b.2 Construct a histogram based on the following data below.

Table 1.11

Class Frequency

0 - 4 2

5 - 9 3

10 - 14 8

15 - 19 7

20 - 24 5

25 - 29 1


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1b.3 Construct a polygon frequency in table 1.12.

Table 1.12

Class Frequency

51 - 55 2

5660 5

61 - 65 9

6670 20

71 - 75 17

76 - 80 7

81 - 85 4

86 - 90 2

91 - 95 0

96 - 100 1


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ANSWERS 1b

1b.1

Cumulative frequency

weight (pound)

5

10

15

20

25

95.5 131.5 185.5149.5 167.5133.5

Less than ogive

More than ogive

125 140


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1b.2

1b.3

0

1

2

3

4

5

6

7

8

-0.5-4.5 4.5-9.5 9.5-14.5 14.5-19.5 19.5-24.5 24.5-29.5

0

5

10

15

20

25

48 53 63 68 73 78 83 88 93 98 103

Midpoint

Frequency

Frequency

Class

Boundary


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PRACTICES

1.a The following data shows the daily expenses for 60 families in buying vegetables in Ipoh, Perak.

Construct a frequency distribution table for the data.

0.97 0.71 1.05 0.78 1.52 1.37

1.24 1.76 0.77 0.88 0.91 1.02

0.78 0.84 0.91 0.93 1.12 1.25

1.26 0.86 0.97 0.74 0.51 1.37

0.33 1.18 0.21 1.62 0.41 0.47

0.69 0.88 1.47 1.02 1.77 0.69

0.51 0.57 1.08 1.51 1.32 0.66

0.73 0.51 1.27 1.16 1.78 0.63

1.61 0.47 1.57 1.26 1.43 0.67

0.46 0.36 1.16 1.96 2.12 1.82

1.b The total export in West Malaysia in million ringgit from 1971 until 1975 are shown below.

From the data, construct a horizontal bar graph.

Year Total

1971

1972

1973

1974

1975

2.640

2.481

3.658

5.221

4.073


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1.c Construct a histogram using the set of data below by starting with class interval : 30-32

30 50 46 42 33 50 43 51 38 50 56 36 36

44 51 55 40 46 48 34 51 46 40 30 35

48 52 31 46 37 41 50 34 54 42 34 32

1.d Construct an ogive using the data below.

Class Frequency

12 2

13 5

14 8

15 6

16 3

17 1


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ANSWERS

1.a

Expenses Number of families

0.20 - 0.40 4

0.40 - 0.60 8

0.60 - 0.80 10

0.80 - 1.00 10

1.00 - 1.20 8

1.20 - 1.40 7

1.40 - 1.60 5

1.60 - 1.80 5

1.80 - 2.00 2

2.00 - 2.20 1

Jumlah 60

1.b

0 2 4 6

1971

1972

1973

1974

1975

million (ringgit)

year


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1.c Histogram

1.d Less than ogive

0

1

2

3

4

5

6

29.5 -

32.5

32.5 -

35.5

35.5 -

38.5

38.5 -

41.5

41.5 -

44.5

44.5 -

47.5

47.5 -

50.5

50.5 -

53.5

53.5 -

56.5

0

5

10

15

20

25

30

11.5 12.5 13.5 14.5 15.5 16.5 17.5

Frequency

Class

Boundary

Cumulative

Frequency

Class

Boundary

ba201 engineering mathematic unit1 - statistics

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