Chapter 4 Measure of Dispersion
8
CHAPTER-IV Measures of Dispersion The measure of dispersion or variations is the measure of the extent of variation or deviation of individual values from the central value. This measure of variation gives a pries idea as to the extent of representative ness of the central value. Definition of Measures of Dispersion Dispersion or spread is the degree of the scatter of variation of the variable’s about a central value. Dispersion is the measure of the variation of the item’s. Dispersion is the mean of extent to which the individual. Range: The range is the absolute difference between the largest and the smallest values in a set of data. Rang= The largest value - The smallest value Example: The following figures are the weights of 12 persons. Calculate the range of the data. 120, 125, 118, 127, 140, 135, 107, 102, 98, 118, 109, 115 Here the highest value is 140 pound, and the lowest value is 98 pound. Therefore, the range is (140-98)= 42 pound. Range calculate from grouped data: Frequency distribution of 250 houses according to their money value Value of house in Tk. Number of houses 100 – 500 25 500 – 1000 42 1000 – 2000 69 2000 – 3000 72 3000 – 4000 21 4000 – 5000 13 5000 – 10000 8 250 In this case 100 is taken as the lowest value and 10,000 as the highest value of a house. Therefore, the range of the values of houses is estimated as (10,000-100) = 9,9000. \ Range is Tk. 9,900. Coefficient of Range = The Mean Deviation : The mean deviation is the mean difference of the items in a sot of data from their average. The formula of mean deviation from mean for ungrouped data. Mean Deviation = Where = The mean X = The value of individual items, N = The number of items. Illustration : 1
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Chapter 4 Measure of Dispersion
Transcript of Chapter 4 Measure of Dispersion
Measures of DispersionCHAPTER-IV
Measures of Dispersion
The measure of dispersion or variations is the measure of the extent of variation or deviation of individual values from the central value. This measure of variation gives a pries idea as to the extent of representative ness of the central value.
Definition of Measures of Dispersion
Dispersion or spread is the degree of the scatter of variation of the variables about a central value.
Dispersion is the measure of the variation of the items.
Dispersion is the mean of extent to which the individual.
Range:
The range is the absolute difference between the largest and the smallest values in a set of data.
Rang= The largest value - The smallest value
Example: The following figures are the weights of 12 persons. Calculate the range of the data.
120, 125, 118, 127, 140, 135, 107, 102, 98, 118, 109, 115
Here the highest value is 140 pound, and the lowest value is 98 pound. Therefore, the range is (140-98)= 42 pound.
Range calculate from grouped data:
Frequency distribution of 250 houses according to their money value
Value of house in Tk.
Number of houses
100 500
25
500 1000
42
1000 2000
69
2000 3000
72
3000 4000
21
4000 5000
13
5000 10000
8
250
In this case 100 is taken as the lowest value and 10,000 as the highest value of a house. Therefore, the range of the values of houses is estimated as (10,000-100) = 9,9000. ( Range is Tk. 9,900.
Coefficient of Range =
100
Value
Smallest
Value
Largest
Range
+
The Mean Deviation:
The mean deviation is the mean difference of the items in a sot of data from their average.
The formula of mean deviation from mean for ungrouped data.
Mean Deviation =
N
x
X
-
Where
X
= The mean
X = The value of individual items, N = The number of items.
Illustration:
The heights of 12 persons are given below in inches, calculation of mean deviation of the heights. 58, 59, 65, 62, 69, 67, 63, 61, 62, 60, 63, 62.
Calculation:
Height in inches (X)
X
X
-
(Ignoring sign)
58
4.58
=
S
=
N
X
Mean
= 62.58 inches.
=
=
S
=
12
16
.
28
N
X
-
X
deviation
Mean
2.35 inches.
59
3.58
60
2.58
61
1.58
62
.58
62
.58
62
.58
63
.42
63
.42
65
2.42
67
4.42
69
6.42
(X=751
(
X
X
-
= 28.16
Calculation of mean deviation from grouped data:
The formula is: MD =
N
X
X
f
-
S
Where, f = The frequency of the class
x = The mid-value of the class
x
= The arithmetic mean
N = The total frequency
X
X
-
= The deviation of the mid-value from the mean disregarding (+ or - ) sign.
Illustration: Computation of Mean Deviation from grouped Data.
Age in years
No. of persons
f
Mid value
X
d=
C
A
X
-
fd
X
X
-
f
X
X
-
20 30
3
25
- 3
- 9
29.72
89.16
30 40
61
35
- 2
- 122
19.72
1202.92
40 50
132
45
- 1
- 132
9.72
1283.04
50 60
153
55=A
0
0
.28
42.84
60 70
140
65
1
140
10.28
1439.20
70 80
51
75
2
102
20.28
1034.28
80 90
2
85
3
6
30.28
60.56
Total
542
(fd=15
00
.
5152
=
-
S
X
X
f
Mean
72
.
54
28
.
0
55
10
542
15
55
=
-
=
-
+
=
S
+
=
C
N
fd
A
X
years
( Mean Deviation =
51
.
9
542
00
.
5152
=
=
-
S
N
X
X
f
years
Calculation of Mean deviation from Median:
Formula: MD =
N
median
X
-
S
Where, X = The individual values
N= No. of items
MD = The mean deviation
(X-median(=The absolute deviation of each value from the median.
Illustration : Daily production records of a factory are given below in thousand tons calculated of the mean deviation from the median for this series.
25, 31, 28, 35, 36, 33, 26, 45, 42, 32, 29, 38.
Table for calculation of mean deviation from 0 median.
Production in a
sending order
[X-Median]
25
7.5
Median =
th
th
N
N
7
1
2
12
6
2
12
2
1
2
2
=
+
+
=
=
+
+
So the mean value of 6th & 7th items is the median
i. e,
5
.
32
2
33
32
=
+
tons
M.D =
12
58
=
-
S
N
median
X
= 4.83 thousand tons.
26
6.5
28
4.5
29
3.5
31
1.5
6th 32
.5
7th 32
.5
35
2.5
36
3.5
38
5.5
42
9.5
45
12.5
N=12
58
Coefficient of mean deviation :
100
mode
or
median
or
mean
deviation
mean
The Standard Deviation
In case of mean deviation we have deliberately disregarded signs to find out the sum of the deviations. This process of ignoring sign is not proper from algebraic point of view. The standard deviation provides a basis for overcoming this mathematical error and at the same time enables us to have a precise measure of dispersion. The deviations of the values from the mean in the ease of standard deviation are not taken in original form but are squared up. Though the process of squaring the deviations the negative character of the deviations is removed and the problem of tie does not arise. The squared deviations are then summed up and their average gives us the variance. The square rout of the variance is the standard deviation.
Calculation of Standard Deviation from Ungrouped Data
Direct Method : The formula is :
(
)
N
X
X
2
-
S
=
s
Where, ( = The standard deviation
x = The individual values
x
= The Arithmetic mean
N= Number of items.
Illustration: Marks secured out of 100 by 10 students of a class are recorded as below. Calculation of standard deviation of the mark secured. 36, 41, 55, 42, 26, 37, 33, 52, 67, 75
Solution:
Marks
X
X-
X
(
)
2
X
X
-
26
- 204
416.16
AM =
4
.
46
10
464
=
Standard Deviation=
(
)
N
X
X
2
-
S
=
s
84
.
218
10
40
.
2188
=
=
s
(= 14.79 marks
33
- 13.4
179.56
36
- 10.4
108.16
37
- 9.4
88.36
41
- 5.4
29.16
42
- 4.4
19.36
52
5.6
31.36
55
8.6
73.96
67
20.6
424.36
75
28.6
817.96
464
2188.40
* Standard deviation by short cut method
Formula -
(
)
2
2
-
-
-
S
=
N
X
X
N
X
X
s
Where, ( = Standard deviation
A = The assumed mean
X = The value of individual items
N = The number of items.
Illustration : The annual income figure at 11 families are recorded below in taka. Calculate the standard deviation by using short cut method. The income figures are:
4500, 5000, 53000, 6500, 5500, 6200, 5700, 6900, 7600, 5275, 5750.
Solution
Annual income in Taka (X)
x-A=D
(X-A)2=D2
4500
- 1200
14,40,000
5000
- 700
490000
5275
- 425
180625
5300
- 400
160000
5500
- 200
40000
5700
0
0
5750
50
2500
6200
500
250000
6500
800
640000
6900
1200
1440000
7600
1900
3610000
Total
((X-A)=1525
(D=(X-A)=1525
((X-A)2=8,253,125
(D2=(X-A)2=8,253,125
SD =
-
2
2
N
d
N
d
S, D =
(
)
(
)
2
2
-
S
-
-
S
=
N
A
X
N
A
X
s
=
2
11
1525
11
125
,
253
,
8
-
=
(
)
2
64
.
138
09
.
750284
-
=
4
.
731063
= 855.02 Taka
( Standard Deviation = 855.02 Taka.
Standard Deviation from Grouped Data
a) Direct Method : The formula
Standard Deviation :
(
)
N
X
X
f
2
-
S
=
s
Where, X = The mid-value of the class
f = The frequency of the class
X
= The mean
N = The total frequency
(
)
2
X
X
-
= Square at the deviation of X from
X
for each class.
Illustration : The following table shows the age of 200 persons. Calculate the standard deviation from the age group in years.
Solution
Age group in years
Number of persons
0 4
37
5 9
38
10 19
33
20 35
43
36 45
20
46 59
18
60 75
11
Age group
f
x (mid value)
fx
x
x
-
(
)
2
x
x
-
f
(
)
2
x
x
-
0 4
37
2.0
74.0
- 20.49
419.84
15534.08
5 9
38
7.0
266.0
- 15.49
239.94
9117.72
1019
33
14.5
478.5
- 7.99
63.84
2106.72
2035
43
27.5
1182.5
5.01
25.10
1079.30
36- 45
20
40.5
810.0
18.01
324.36
6487.20
4659
18
52.5
945.0
30.01
900.60
16210.80
6075
11
67.5
742.5
45.01
2025.90
22284.90
20
4498.5
72820.72
Mean,
49
.
22
200
5
.
4498
=
=
S
=
N
fx
X
years
Standard deviation =
(
)
200
72
.
72820
2
=
-
S
N
X
X
f
=
1036
.
64
.
3
= 19.08 years.
b) Standard deviation from grouped data by using s
c) hort-cut method:
The formula:
Standard Deviation =
C
N
fd
N
fd
-
=
2
2
s
Where, C = the size of class interval
f = The class frequency
d =
C
A
X
-
X = The mid value of individual class
A = The assumed mean
N = The total frequency.
Illustration : Calculate the standard deviation of the following distribution of the weights at a group of 1000 children:
Solution
Weight groups
Frequency
24.5 28.5
3
28.5 32.5
11
32.5 36.5
61
36.5 40.5
147
40.5 44.5
258
44.5 48.5
276
48.5 52.5
135
52.5 56.5
73
56.5 60.5
32
60.5 64.5
4
Table for calculation of standard deviation by short-cut method:
Weight group
Frequency (f)
Mid-value (X)
Step deviation
C
A
X
d
-
=
d2
fd
fd2
24.5 28.5
3
26.5
- 4
16
- 12
48
28.5- 32.5
11
30.5
- 3
9
- 33
99
32.5-36.5
61
34.5
- 2
4
- 122
244
36.5-40.5
147
38.5
- 1
1
- 147
147
40.5-44.5
258
42.5=A
0
0
0
0
44.5-48.5
276
46.5
1
1
276
276
48.5-52.5
135
50.5
2
4
270
540
52.5-56.5
73
54.5
3
9
219
657
56.5-60.5
32
58.5
4
16
128
512
60.5-64.5
4
62.5
5
25
20
100
Total
1000
599
2623
Here d =
4
5
.
42
-
X
, c = 4, A = 42.5
C
N
fd
N
fd
S
-
S
=
2
2
s
=
4
1000
599
1000
2623
2
-
=
(
)
4
6
.
623
.
2
2
-
=
4
36
.
623
.
2
-
=
4
263
.
2
= 1.504(4 = 6.016
( Standard Deviation = 6.016
C
N
fd
A
X
+
=
EMBED Equation.3
4
100
599
5
.
42
+
=
= 42.5+5.99 (4 = 42.5+23.96 = 66.46
100
=
X
CV
s
Variance =
2
s
=
100
46
.
66
016
.
6
EMBED Equation.3
Variance
=
s
= 9.05, CV = Coefficient of Variation
100
=
X
s
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