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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
Step 5 : Write the title.
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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
Step 4 : Write the title.
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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