Prepared by : NURZATY BINTI MUHAMAD NOR Page 1 B3001/UNIT3/1 Unit 3 MEASURES OF DISPERSION Understand frequency distribution graph and understand measures of dispersion. On completion of this unit, the students should be able to : 1. Determine quartile, decile and percentile from ogive graph. 2. Calculate mean deviation, variance and standard deviation for ungrouped data. 3. Calculate mean deviation, variance and standard deviation for grouped data. General Objective Specific Objective
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BA201 Engineering Mathematic UNIT3 - Measures of Dispersion
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Prepared by : NURZATY BINTI MUHAMAD NOR Page 1
B3001/UNIT3/1
Unit
3
MEASURES OF
DISPERSION
Understand frequency distribution graph
and understand measures of dispersion.
On completion of this unit, the students
should be able to :
1. Determine quartile, decile and
percentile from ogive graph.
2. Calculate mean deviation,
variance and standard deviation
for ungrouped data.
3. Calculate mean deviation,
variance and standard deviation
for grouped data.
General Objective
Specific Objective
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Decile (D)
Percentile
(P)
3.0 INTRODUCTION
Measures of dispersion are the measures of the degree of dispersion to
determine how far the values of data in a set of data scatter or spread out from its
average value. There are a few types of measures of dispersion :
a) Quartile
b) Decile
c) Percentile
d) Mean Deviation and Variance
e) Standard deviation
What is the Quartile,Decile and Percentile in
measures of dispersion?
Quartile (Q)
Value which divides a set of data arranged in
ascending or decending order into four equal parts.
There are first quartile (Q1), second quartile /
median (Q2) and third quartile (Q3)
Value which divides a set of data arranged in ascending or
decending order into ten equal parts. These are known as
D1, D2, D3,……,D9.
Value which divides a set of data arranged in ascending or
decending order into hundred equal part. These are known
as P1, P2, P3,……,P99.
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3.1 MEASURES OF DISPERSION
Measures of dispersion are the measures of the degree of dispersion to
determine how far the values of data in a set of data scatter or spread out from its
average value. There are a few types of measures of dispersion :
f) Quartile
g) Decile
h) Percentile
i) Variance
j) Standard deviation
3.1.1 Quartile,Decile and Percentile
Firstly, from data given , we must construct a cumulative frequency graph of
100% against class boundaries of data to determine quartile,decile and percentile
using ogive as shown below:
Example 3.1:
From data in table 3.1, determine first quartile ,third quartile fourth decile and eighty
eighth percentile through ogive.
Table 3.1
Order (RM) No. of order
10 and less than 20 85
20 and less than 30 120
30 and less than 40 225
40 and less than 50 135
50 and less than 60 105
60 and less than 70 30
INPUT
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Solution :
Step 1 : Construct cumulative frequency table as shown in Table 3.2.
Jadual 3.2
Order (RM) Cumulative
frequency (cf)
% cumulative
frequency (% cf)
less than 10 0 0
less than 20 85 12.1
less than 30 205 29.3
less than 40 430 61.4
less than 50 565 80.7
less than 60 670 95.7
less than 70 700 100.0
Step 2 : Construct ogive graph
From the graph 3.1,
Figure 3.1: Ogive Graph ‘Less Than’
(85/100) x 700
= 12.1 %
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First quartile, K1 = RM 27.00
Third quartile, K2 = RM 47.00
Fourth decile, D4 = RM 33.10
Eighty eighth Percentile , P88 = RM 54.50
ACTIVITY 3a
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3a.1 A set of grouped data shows the weight of 100 students. Find first quartile,
third quartile,fifth decile and sixtieth percentile,P60.
Berat (kg) Kekerapan
45 – 48 4
49 – 52 7
53 – 56 10
57 – 60 14
61 – 64 19
65 – 68 25
69 - 72 12
73 - 76 9
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FEEDBACK 3a
3a.1 first quartile, K1 = 57.7 kg
third quartile, K3 = 67.7 kg
fifth decile, D5 = 64.0 kg
sixtieth percentile P60 = 67.0 kg
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3.2 MEAN DEVIATION FOR UNGROUPED DATA
Mean deviation for the raw data x1, x2, x3,……………………..xn is calculated by
using the following formula.
E =
n
xxn
li
i
……………………………(3.1)
where x = mean.
Example 3.2:
Calculate mean deviation for the following data.
a) 12, 6, 15, 3, 12, 6, 21, 15, 18, 12
b) 12, 12, 12, 12, 12, 12, 12, 12, 12, 12
Solution:
a)
12 6 15 3 12 6 21 15 18 12 120
xxi 0 -6 3 -9 0 6 9 3 6 0
xxi 0 -6 3 -9 0 6 9 3 6 0 4.2
E = n
xxi =
10
42 = 4.2
b)
12 12 12 12 12 12 12 12 12 12 120
xxi 0 0 0 0 0 0 0 0 0 0
xxi 0 0 0 0 0 0 0 0 0 0 0
E = n
xxi =
10
0 = 0
INPUT
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Example 3.3:
Determine mean deviation for the following data.
5, 7, 1, 2, 4
Solution:
ix 5 7 1 2 4 19
xxi 1.2 3.2 -2.8 -1.8 0.2
xxi 1.2 3.2 2.8 1.8 0.2 9.2
Mean x = n
xn
li
i =
5
19 = 3.8
mean deviation, E:
E = n
xxn
li
i
= 5
2.9 = 1.84
3.3 MEAN DEVIATION FOR GROUPED DATA
Mean deviation for grouped data, E is calculated by using the following
E = n
fxxk
li
ii
………………………………. (3.2)
Where x = mean .
Example 3.4:
Calculate mean deviation for the following grouped data.
Class Frequency if
0 - 4
5 - 9
10 - 14
15 - 19
30
51
10
10
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Solution:
ix if ii fx xxi xxi
ii fxx
2
7
12
17
30
51
10
10
60
357
120
170
-5
0
5
10
5
0
5
10
150
0
50
100
707 300
mean, x = n
fx ii =
101
707 = 7
mean deviation, E = n
fxx ii =
101
300 = 3
Example 3.5:
Calculate mean deviation for the following grouped data.
Class Frequency, if
24 – 32
34 – 42
44 – 52
54 – 62
64 – 72
5
8
12
14
11
50
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Solution:
Mid class, ix if ii fx xxi xxi ii fxx
28
38
48
58
68
5
8
12
14
11
140
304
576
812
748
-23.6
-13.6
-3.6
6.4
16.4
23.6
13.6
3.6
6.4
16.4
118.0
108.0
43.2
89.6
180.4
50 2580 540
Mean = n
fxk
li
ii =
50
2580 = 51.6
And mean deviation = n
fxxk
li
ii
= 50
540
= 10.8
(24 +32) / 2
= 28
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ACTIVITY 3b
3b.1 Calculate mean deviation for the following data.:
Class Frequency if
0 – 4
4 – 8
8 – 12
12 – 16
16 – 20
4
16
20
16
4
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ANSWERS 3b
3b.1 Mean, E = 3.2
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3.3 VARIANCE FOR UNGROUPED DATA
Variance (S2
) for the raw data x1, x2, x3, ………xn is calculated by using the
following formula :
S2 =
1
2
n
xxn
li
i
…………………………………………(3.3)
Where x = mean.
Example 3.6:
Calculate the variance for the following data :
a) 12, 6, 15, 3, 12, 6, 21, 15, 18, 12
b) 12, 10, 12, 14, 10, 13, 12, 11, 14, 12
c) 12, 12, 12, 12, 12, 12, 12, 12, 12, 12
Solution:
a)
12 6 15 3 12 6 21 15 18 12 120
xxi 0 -6 3 -9 0 -6 9 3 6 0
2xxi 0 36 9 81 0 36 81 9 36 0 288
S2 =
1
2
n
xxi =
110
288
= 32
INPUT
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b)
12 10 12 14 10 13 12 11 14 12 120
xxi 0 -2 0 2 -2 1 0 -1 2 0
2xxi 0 36 9 81 0 36 81 9 36 0 18
S2 =
1
2
n
xxi =
110
18
= 2
c)
12 12 12 12 12 12 12 12 12 12 120
xxi 0 0 0 0 0 0 0 0 0 0
2xxi 0 0 0 0 0 0 0 0 0 0 0
S2 =
1
2
n
xxi =
110
0
= 0
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ACTIVITY 3c
3c.1 Find the mean and variance for the following data:
10, 7, 19, 13, 14, 9
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FEEDBACK 3c
3c.1 mean = 12
variance = 15.3
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3.4 STANDARD DEVIATION
Standard deviation is represented as S, square root of variance.
S= iancevar
Example 3.7:
Determine standard deviation for data in (a), (b) and (c) in example 3.6.
Solution:
Standard deviation :
(a) S2 = 32 , S = 32 = 5.66
(b) S2 = 2 , S = 2 = 1.41
(c) S2 = 0 , S = 0 = 0
Variance Formula for Ungrouped Data
You can use the following formula to find the variance for ungrouped data :
S2 =
1
2
n
xxn
li
i
………………………………(3.4)
Or
S2 =
1
22
n
xnxn
li
i
………………………………(3.5)
where x = mean.
INPUT
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Example 3.8:
Determine a variance of the following using data : 5, 7, 1, 2, 4.
Solution :
ix xxi 2xxi 2
xx
5
7
1
2
4
1.2
3.2
-2.8
-1.8
-0.2
1.44
10.44
7.84
3.24
0.04
25
46
1
4
6
19 22.8 95
mean,
x = 5
ix =
5
19 =3.8
First method – Using equation (3.4):
S2 =
1
2
n
xxn
li
i
= 4
8.22 = 5.7
Second method – Using equation (3.5):
S2 =
1
22
n
xnxn
li
i
=4
)8.3(595 2
= 4
2.7295
= 4
8.22
= 5.7
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Variance Formula for Grouped Data
You can use the following formula to find the variance for grouped data :
1
2
2
n
fxx
S
k
li
ii
………………………………(3.6)
Or
S2 =
1
22
n
xnfxk
li
ii
………………………………(3.7)
where x = mean.
Example 3.9:
Calculate the variance and standard deviation for the following distribution
frequency table using the formula in equation 3.6.
Class Frequency
0 – 4
5 – 9
10 – 14
15 – 19
30
51
10
10
Solution:
Class ix if ix if xxi 2xxi 2xxi if
0 – 4
5 – 9
10 – 14
2
7
12
30
51
10
60
357
120
-5
0
5
25
0
25
750
0
250
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15 – 19 17 10 170 10 100 1000
So,
x = 101
707 = 7
and
Variance,
1
2
2
n
fxx
S
k
li
ii
= 1101
2000
= 20
standard deviation, S = 20 = 4.47
Example 3.10:
Using formula in equation 3.7, calculate the variance and standard deviation in
the following table.
Class Frequency
0 – 4
5 – 9
10 – 14
15 – 19
30
51
10
10
Solution:
Class ix 2
ix if ix if 2
ix if
0 – 4
5 – 9
10 – 14
15 – 19
2
7
12
17
4
49
144
289
30
51
10
10
60
357
120
170
120
2499
1440
2890
101 707 6949
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7101
707
n
fxx
ii
2S =1
22
n
xnfxk
li
ii
=
100
71016949 7
20100
1000
100
49496949
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ACTIVITY 3d
3d.1 Calculate the mean and standard deviation for the following frequency
distribution table :
Class Frequency
16 – 21
22 – 27
28 – 33
24 – 39
15
16
5
5
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FEEDBACK 3d
3d.1 Mean, 5.24
x
Standard deviation, S = 6
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PRACTICES
3.a Find mean, mean deviation and standard deviation using the following data.