Measures of Dispersion: Absolute and Relative Measures of Dispersion, their computation & properties.

3. Definition of Measures of Dispersion:

7. The four important absolute measures of dispersion are as follows:
i. Range
ii. Mean or average deviation
iii.Standard deviation
iv.Quartile deviation

8. When the data is in different units, in such a situation we may use the relative
dispersion. A relative dispersion is independent of original units. Generally, relative
measures of dispersion are expressed in terms of ratio, percentage etc.
The relative measures of dispersion are as follows:
1. Coefficient of range
2. Coefficient of mean deviation
3. Coefficient of variation
4. Coefficient of quartile deviation

9. The range of a set of observation is the difference between two extreme values, i.e the
difference between the maximum and minimum values.
Therefore, it indicates the limits within all observations fall.
In the form of an equation:
Range = Highest value – Lowest value

10. Let us consider a set of observations 𝑥1, 𝑥2,𝑥3, ……………… , 𝑥 𝑛 and 𝑋 𝐻 is
maximum and 𝑋 𝐿 is minimum.
Then Range = 𝑋 𝐻 − 𝑋 𝐿.
Example:
Find out the range of the set of observations, -7, -2, -4, 0, 8.
Solution:
Here, maximum value, 𝑋 𝐻 = 8 and minimum value, 𝑋 𝐿 = −7
Range = 𝑋 𝐻 − 𝑋 𝐿 = 8 − −7 = 8 + 7 = 15

11. Range For Grouped data:
In this case, the range is the difference between the upper boundary of the highest
class and the lower boundary of the lowest class.
Then Range = 𝑋 𝑈 − 𝑋 𝐿
Where,
𝑋 𝑈= The upper boundary of the highest class.
𝑋 𝐿= The lowest boundary of the highest class.

12. Example: determine the range from the following frequency distribution.
Salary (TK.) 1700-1800 1800-1900 1900-2000 2000-2100 2100-2200
No. of workers 420 460 500 300 200
Solution:
From the given frequency distribution,
We have,
The upper boundary of the highest class, 𝑋 𝑈 = 𝑇𝐾. 2200
And the lowest boundary of the highest class, 𝑋 𝐿 = 𝑇𝐾. 1700
Then Range = 𝑋 𝑈 − 𝑋 𝐿 = 𝑇𝐾. 2200 − 𝑇𝐾. 1700 = 𝑇𝐾. 500

13. 1. It is the simplest measure of dispersion.
2. It is easy to understand and calculate the range.
3. The important merit of range is that it gives us a quick idea of the variability of a set of
data.
4. It does not depend on the measures of central tendency.
Advantages of Range:

14. 1. It is influenced by the extreme values.
2. It cannot be computed for open end distribution.
3. It is no suitable for further mathematical treatment.
Disadvantages of Range:

15. 1. It is time saving and widely used in industrial quality control, weather forecast.
2. Variations in stock exchange can be studies by range.
Use of Range:

16. Mean deviation or Average deviation:
Definition of Mean deviation:
Mean deviation is the mean of absolute deviations of the items from an average like
mean, median or mode. Normally, we consider the arithmetic mean as the average.

17. 1. Mean deviation for ungrouped data:
If 𝑥1, 𝑥2,𝑥3, ……………… , 𝑥 𝑛 be a set of n observations or values, then the mean
deviation is expressed and defined as:
i. Mean deviation about arithmetic mean, M.D (𝑋) =
𝑥 𝑖;𝑥𝑛
𝑖=1
𝑛
; 𝑋 = 𝐴𝑟𝑖𝑡𝑕𝑚𝑒𝑡𝑖𝑐 𝑚𝑒𝑎𝑛
ii. Mean deviation about median, M.D (𝑋) =
𝑥 𝑖;𝑀𝑒𝑛
𝑖=1
𝑛
; 𝑀𝑒 = 𝑀𝑒𝑑𝑖𝑎𝑛
iii.Mean deviation about mode, M.D (𝑋) =
𝑥 𝑖;𝑀𝑜𝑛
𝑖=1
𝑛
; 𝑀𝑜 = 𝑀𝑜𝑑𝑒

18. Example:
Find out the mean deviation from the following given set 2,3,4,5,6.
Solution: We know,
Mean deviation, M.D (𝑋) =
𝑥 𝑖;𝑥𝑛
𝑖=1
𝑛
Here, 𝑋 =
𝑥 𝑖
𝑛
𝑖=1
𝑛
=
2:3:4:5:6
5
= 4
Mean deviation, M.D (𝑋) =
2;4 : 3;4 : 4;4 : 5;4 : 6;4
5
=
6
5
= 1.2

19. Example:
The number of patients seen in the emergency room at Ibrahim Memorial Hospital for a
sample of 5 days last year were: 103, 97, 101, 106 and 103.
Determine the mean deviation and interpret.

20. Solution: Table for calculation of Mean Deviation
No. of patients (x) 𝑥 − 𝑥 = 𝑥 − 102
103 𝑥 − 102 = 103 − 102 = 1
97 𝑥 − 102 = 97 − 102 = 5
101 1
106 4
103 1
𝑥𝑖 = 510 𝑥𝑖 − 𝑥 = 12
Arithmetic Mean,
𝑥 =
𝑥𝑖
𝑛
𝑖<1
𝑛
=
510
5
= 102
Mean Deviation about mean, M.D (𝑋) =
𝑥 𝑖;𝑥𝑛
𝑖=1
𝑛
=
12
5
= 2.4
Interpretation:
The mean deviation is 2.4 patients per day. The no. of patients deviates, on average by 2.4
patients from the mean of 102 patients per day.

21. 1. Mean deviation for grouped data:
If 𝑥1, 𝑥2,𝑥3, ……………… , 𝑥 𝑛 occur with frequencies 𝑓1, 𝑓2,𝑓3, ……………… , 𝑓𝑛
respectively then the mean deviation can be written as:
i. Mean deviation about arithmetic mean, M.D (𝑋) =
𝑓 𝑖 𝑥 𝑖;𝑥𝑛
𝑖=1
𝑛
ii. Mean deviation about median, M.D (𝑋) =
𝑓 𝑖 𝑥 𝑖;𝑀𝑒𝑛
𝑖=1
𝑛
iii.Mean deviation about mode, M.D (𝑋) =
𝑓 𝑖 𝑥 𝑖;𝑀𝑜𝑛
𝑖=1
𝑛

22. Example:
Calculate mean deviation from the following data
Income (TK.) 0-10 10-20 20-30 30-40 40-50 50-60 60-70
No. of Persons 6 8 10 12 7 4 3

23. Solution:
Table for calculation of Mean Deviation from mean
Income (TK.) Class mid-
point (x)
No. of persons
(f)
𝒇𝒙 𝒙 − 𝒙 𝒇 𝒙 − 𝒙
0-10 5 6 30 26 156
10-20 15 8 120 16 128
20-30 25 10 250 6 60
30-40 35 12 420 4 48
40-50 45 7 315 14 98
50-60 55 4 220 24 96
60-70 65 3 195 34 102
Total 𝑓𝑖 = 50 𝑓𝑖 𝑥𝑖 = 1550 𝑓𝑖 𝑥𝑖 − 𝑥 = 688

24. We know, Arithmetic Mean,
𝑥 =
𝑓𝑖 𝑥𝑖
𝑛
𝑖<1
𝑛
=
1550
50
= 31
Therefore, Mean deviation about arithmetic mean,
M.D (𝑋) =
𝑓 𝑖 𝑥 𝑖;𝑥𝑛
𝑖=1
𝑛
=
688
50
= 13.76

25. 1. It is easy to calculate and understand.
2. It is based on all the observation.
3. It is useful measure of dispersion.
4. It is not greatly affected by extreme values.
Advantages of Mean Deviation:

26. 1. It is not suitable for further mathematical treatment.
2. Algebraic positive and negative signs are ignored.
Disadvantages of Mean Deviation:
Uses of mean deviation:
1. It is used in certain economic and anthropological studies.

27. Definition:
The standard deviation is the positive square root of the mean of the squared deviation from
their mean of a set of observation. It can be written as
Standard Deviation:
The standard deviation is the most important measure of dispersion.
Standard Deviation, 𝜎 =
𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑜𝑓 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑓𝑟𝑜𝑚 𝑚𝑒𝑎𝑛
𝑇𝑜𝑡𝑎𝑙 𝑛𝑜. 𝑜𝑓 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛

28. 1. Standard Deviation for Ungrouped Data:
Let,𝑥1, 𝑥2,𝑥3, ……………… , 𝑥 𝑛 be a set of n observations or values, then their
standard deviation can be written as,
Standard Deviation, 𝜎𝑥 =
(𝑥 𝑖
𝑛
𝑖=1 ;𝑥)2
𝑛
=
𝑥 𝑖
2
𝑛
− (
𝑥 𝑖
𝑛
)2

29. 2. Standard Deviation for Grouped Data:
If 𝑥1, 𝑥2,𝑥3, ……………… , 𝑥 𝑛 occur with frequencies 𝑓1, 𝑓2,𝑓3, ……………… , 𝑓𝑛
respectively then the standard deviation can be written as:
Standard Deviation, 𝜎𝑥 =
𝑓 𝑖(𝑥 𝑖
𝑛
𝑖=1 ;𝑥)2
𝑛
=
𝑓 𝑖 𝑥 𝑖
2
𝑛
− (
𝑓 𝑖 𝑥 𝑖
𝑛
)2

30. Example:
Find the standard deviation for the following data.
75, 73, 70, 77, 72, 75, 76, 72, 74, 76

31. Values,(X) (𝑥 − 𝑥) (𝑥 − 𝑥)2
75 1 1
73 -1 1
70 -4 16
77 3 9
72 -2 4
75 1 1
76 2 4
72 -2 4
74 0 0
76 2 4
𝑥𝑖 = 740 (𝑥 − 𝑥)2 = 592
Solution:
Table for calculation of standard deviation

32. We know,
Arithmetic mean, 𝑥 =
𝑥 𝑖
𝑛
𝑖=1
𝑛
=
740
10
= 74
Standard Deviation, 𝜎𝑥 =
(𝑥 𝑖
𝑛
𝑖=1 ;𝑥)2
𝑛
=
44
10
= 2.09

33. Example:
Calculate standard deviation from the following data
Income (TK.) 0-10 10-20 20-30 30-40 40-50 50-60 60-70
No. of Persons 6 8 10 12 7 4 3

34. Solution:
Table for calculation of standard deviation
Income (TK.) Class mid-
point (x)
No. of
persons (f)
𝒇𝒙 (𝒙 − 𝒙) 𝟐
𝒇(𝒙 − 𝒙) 𝟐
0-10
5 6 30
(𝒙 − 𝒙) 𝟐
= (𝟓 − 𝟑𝟏) 𝟐
= 𝟔𝟕𝟔
𝒇(𝒙 − 𝒙) 𝟐
= 𝟔 × 𝟔𝟕𝟔
= 𝟒𝟎𝟓𝟔
10-20 15 8 120 256 2048
20-30 25 10 250 36 360
30-40 35 12 420 16 192
40-50 45 7 315 196 1372
50-60 55 4 220 576 2304
60-70 65 3 195 1156 3468
Total 𝑓𝑖 = 50 𝑓𝑖 𝑥𝑖 = 1550 𝑓𝑖(𝒙𝒊 − 𝒙) 𝟐
= 13800

35. We know,
Arithmetic Mean, 𝑥 =
𝑓 𝑖 𝑥 𝑖
𝑛
𝑖=1
𝑛
=
1550
50
= 31
Therefore, standard deviation, 𝜎𝑥 =
𝑓 𝑖(𝒙 𝒊;𝒙) 𝟐
𝑛
=
13800
50
= 16.61

36. 1. It is rigidly defined.
2. It is less affected by sampling fluctuations.
3. It is useful for calculating the skewness, kurtosis,
coefficient of correlation, coefficient of variation and so on.
4. It measure the consistency of data.
Advantages of Standard deviation:

37. 1. It is not so easy to compute.
2. It is affected by extreme values.
Disadvantages of standard deviation
1. It is useful for calculating the skewness, kurtosis,
coefficient of correlation, coefficient of variation
and so on.
2. It measure the consistency of data.
Use of standard deviation:

38. Quartile Deviation (Or Semi-interquartile Range):
The quartile deviation is another type of range obtained from the quartiles. It is obtained by
dividing the difference between upper quartile (𝑄3and lower quartile 𝑄1by 2.
Mathematically, the quartile deviation (Q.D) can be written as:
𝑄. 𝐷 =
𝑈𝑝𝑝𝑒𝑟 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒;𝐿𝑜𝑤𝑒𝑟 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒
2
=
𝑄3;𝑄1
2
The term (𝑄3 − 𝑄1) is known as the interquartile range and the quartile deviation
𝑄3;𝑄1
2
is also known as semi-interquartile range.

39. Example:
The Automobile Association checks the prices of gasoline before many holiday weekends.
Listed below are the self-service prices for a sample of 8 retail outlets during the May 2004
Memorial Day Weekend.
40, 22, 60, 30, 45, 66, 70, 55
Determine the quartile deviation, Interquartile range and semi-interquartile range.

40. Solution:
By arranging the given data in ascending order, we have 22, 30, 40, 45, 55, 60, 66, 70.
Here, total number of observation, n = 8 (even)
First quartile, 𝑄1 =
𝑛
4
𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛:(
𝑛
4
:1)𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
2
𝑄1 =
8
4
𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛:(
8
4
:1)𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
2
𝑄1 =
2𝑛𝑑 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛:3𝑟𝑑 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
2
𝑄1 =
30:40
2
= 35

41. Third quartile, 𝑄3 =
3𝑛
4
𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛:(
3𝑛
4
:1)𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
2
𝑄3 =
24
4
𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛:(
24
4
:1)𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
2
𝑄3 =
6𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛:7𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
2
𝑄3 =
60:66
2
= 63
So, Quartile Deviation, (Q.D) =
𝑄3;𝑄1
2
=
63;35
2
= 14
Interquartile range, (IQR) = (𝑄3 − 𝑄1) = 63 − 35 = 28
Semi-interquartile Range =
𝑄3;𝑄1
2
= 14

42. Example:
Find out the quartile deviation, interquartile range and semi-interquartile range from the
following frequency distribution:
Class Less than 10 10-15 15-20 20-25 25-30 More than 30
Frequency 2 6 7 10 3 1

43. Solution:
Table for calculation
Class Freque
ncy (f)
Cumulative
Frequency
(c.f)
Less than 10 2 2
10-15 6 8
15-20 7 15
20-25 10 25
25-30 3 28
More than 30 1 29
Here, Location of first quartile,
𝑄1 =
𝑁
4
𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
=
29
4
𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
= 7.25 𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
Therefore, 7.25 th observation is first quartile
which is lies in the class 10-15.

44. We know,
First quartile, 𝑄1 = 𝐿 +
𝑁
4
;𝐹𝑐
𝑓
× 𝑐
Where,
L= Lower limit of the first quartile class = 10
𝐹𝑐 = 𝐶𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑡𝑕𝑒 𝑝𝑟𝑒𝑐𝑑𝑖𝑛𝑔 𝑡𝑕𝑒 𝑓𝑖𝑟𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒 𝑐𝑙𝑎𝑠𝑠 = 2
𝑓 = 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑡𝑕𝑒 𝑓𝑖𝑟𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒 𝑐𝑙𝑎𝑠𝑠 = 6
𝑁 = 𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛 = 29
𝑐 = 𝐶𝑙𝑎𝑠𝑠 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 = 5
First quartile, 𝑄1 = 10 +
7.25;2
6
× 5 = 14.375

45. Here, Location of first quartile, 𝑄3 =
3𝑁
4
𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
=
87
4
𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
= 21.75 𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
Therefore, 21.75 th observation is third quartile which is lies in the class 20-25.
Third quartile, 𝑄3 = 𝐿 +
3𝑁
4
;𝐹𝑐
𝑓
× 𝑐
𝑄3 = 20 +
21.75;15
10
× 5
= 23.375
Here, L= 20, 𝐹𝑐 = 15,𝑓 = 10, 𝑐 = 5

46. So, Quartile Deviation,
Q.D =
𝑄3;𝑄1
2
=
23.375;14.375
2
= 4.5
Interquartile range,
IQR = (𝑄3 − 𝑄1)
= 23.375 − 14.375 = 9
Semi-interquartile Range =
𝑄3;𝑄1
2
= 4.5

47. Coefficient of Range:
Coefficient of Range, CR=
𝑅𝑎𝑛𝑔𝑒
𝐻𝑖𝑔𝑕𝑒𝑠𝑡 𝑉𝑎𝑙𝑢𝑒 : 𝐿𝑜𝑤𝑒𝑠𝑡 𝑉𝑎𝑙𝑢𝑒
× 100
Coefficient of Range for Ungrouped Data:
Coefficient of Range, CR=
𝑅𝑎𝑛𝑔𝑒
𝑋 𝐻 : 𝑋 𝐿
× 100 ;
=
𝑋 𝐻 − 𝑋 𝐿
𝑋 𝐻 + 𝑋 𝐿
× 100
𝑋 𝐻 = 𝐻𝑖𝑔𝑕𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒,
𝑋 𝐿 = 𝐿𝑜𝑤𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒

48. Example:
Compute the coefficient of range from the following data
10, 12, 8, 5, 6, 20
Solution:
𝑋 𝐻 = 𝐻𝑖𝑔𝑕𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒 = 20,
𝑋 𝐿 = 𝐿𝑜𝑤𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒 = 5
Coefficient of Range, CR =
𝑅𝑎𝑛𝑔𝑒
𝑋 𝐻 : 𝑋 𝐿
× 100 ;
=
20;5
20:5
× 100
= 60%

49. Coefficient of Range for Grouped Data:
Coefficient of Range, CR =
𝑅𝑎𝑛𝑔𝑒
𝑋 𝑈 : 𝑋 𝐿
× 100 ;
=
𝑋 𝑈; 𝑋 𝐿
𝑋 𝑈 : 𝑋 𝐿
× 100
𝑊𝑕𝑒𝑟𝑒,
𝑋 𝑈 = 𝑈𝑝𝑝𝑒𝑟 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑜𝑓 𝑡𝑕𝑒 𝑕𝑖𝑔𝑕𝑒𝑠𝑡 𝑐𝑙𝑎𝑠𝑠
𝑋 𝐿 = 𝐿𝑜𝑤𝑒𝑠𝑡 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑜𝑓 𝑡𝑕𝑒 𝑙𝑜𝑤𝑒𝑠𝑡 𝑐𝑙𝑎𝑠𝑠

50. Example:
Calculate the coefficient of range from the following frequency distribution,
Age (Year) 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 4 7 15 18 16 12 8

51. Solution:
Here, 𝑋 𝑈 = 𝑈𝑝𝑝𝑒𝑟 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑜𝑓 𝑡𝑕𝑒 𝑕𝑖𝑔𝑕𝑒𝑠𝑡 𝑐𝑙𝑎𝑠𝑠 = 80
𝑋 𝐿 = 𝐿𝑜𝑤𝑒𝑠𝑡 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑜𝑓 𝑡𝑕𝑒 𝑙𝑜𝑤𝑒𝑠𝑡 𝑐𝑙𝑎𝑠𝑠 = 10
Coefficient of Range,
CR =
𝑅𝑎𝑛𝑔𝑒
𝑋 𝑈 : 𝑋 𝐿
× 100 ;
=
80;10
80 :10
× 100
= 77.78%

52. Coefficient of Variation (CV):
Coefficient of variation is the most commonly used measure of relative measures of
dispersion. It is 100 times of a ratio of the standard deviation to the arithmetic mean. It is
denoted by C.V. and written as:
𝐶. 𝑉 =
𝜎
𝑥
× 100; 𝑥 ≠ 0

53. Example:
Compute the coefficient of variation from the following data:
Monthly Income 2501-5000 5001-7500 7501-10000 10001-12500 12501-15000
No. of Families 65 130 215 100 70

54. Solution:
Table for calculation of C.V.
Monthly
Income
No. of
Families
Mid value (x) 𝑓𝑥 𝑓𝑥2
2501-5000 65 3750.5 243782.5 914306266.3
5001-7500 130 6250.5 812565 5078937533
7501-10000 215 8750.5 1881358 16462818804
10001-12500 100 11250.5 1125050 12657375025
12501-15000 70 13750.5 962535 13235337518
𝑓 = 580 𝑓𝑥 = 5025290 𝑓𝑥2
= 48348775145

55. Arithmetic Mean, 𝑥 =
𝑓𝑥
𝑓
=
5025290
580
= 8664.29
Standard Deviation, 𝜎 =
𝑓𝑥2
𝑁
− (
𝑓𝑥
𝑁
)2 =
5025290
580
− (
48348775145
580
)2 = 2879.25
Coefficient of Variation, 𝐶. 𝑉 =
𝜎
𝑥
× 100 =
2879.25
8664.29
× 100 = 33.23%

56. Example:
The run-scores of two cricketers for 10 innings are given below:
Cricketers-A 114 45 0 31 75 102 198 8 0 7
Cricketers-B 15 25 18 30 11 4 23 21 31 22
Who of the two is a more consistent batsman?

57. Solution:
In order to find out who batsman is more consistent, we have to calculate the coefficient of
variation for each batsman.
Table for calculation of C.V.
Cricketer-A Cricketer-B
Score (x) 𝒙 𝟐 Score (y) 𝒚 𝟐
114 12996 15 225
45 2025 25 625
0 0 18 324
31 961 30 900
75 5625 11 121
102 10404 4 16
198 39204 23 529
8 64 21 441
0 0 31 961
7 49 22 484
𝑥 = 580 𝑥2
= 71328 𝑦 = 200 𝑦2
= 4626

58. We know, Coefficient of Variation, 𝐶. 𝑉 =
𝜎
𝑥
× 100
For Cricketer-A
Arithmetic Mean,
𝒙 =
𝒙
𝒏
=
𝟓𝟖𝟎
𝟏𝟎
= 𝟓𝟖
Standard Deviation,
𝝈 =
𝒙 𝟐
𝒏
− (
𝒙
𝒏
) 𝟐
=
𝟕𝟏𝟑𝟐𝟖
𝟏𝟎
− (
𝟓𝟖𝟎
𝟏𝟎
) 𝟐
= 𝟔𝟏. 𝟑𝟗
Coefficient of Variation,
𝑪. 𝑽 =
𝝈
𝒙
× 𝟏𝟎𝟎 =
𝟔𝟏. 𝟑𝟗
𝟓𝟖
× 𝟏𝟎𝟎
= 𝟏𝟎𝟓. 𝟖𝟒%
For Cricketer-B
Arithmetic Mean,
𝒚 =
𝒚
𝒏
=
𝟓𝟖𝟎
𝟏𝟎
= 𝟐𝟎
Standard Deviation,
𝝈 =
𝒚 𝟐
𝒏
− (
𝒚
𝒏
) 𝟐
=
𝟒𝟔𝟐𝟔
𝟏𝟎
− (
𝟐𝟎𝟎
𝟏𝟎
) 𝟐
= 𝟕. 𝟗𝟏
Coefficient of Variation,
𝑪. 𝑽 =
𝝈
𝒚
× 𝟏𝟎𝟎 =
𝟕. 𝟗𝟏
𝟐𝟎
× 𝟏𝟎𝟎
= 𝟑𝟗. 𝟓𝟔%
Comment:
From the above result, we see that, C.V. (A) = 105.84% and C.V. (B) 39.56%
Since, C.V. (A)> 𝐂. 𝐕. (𝐁). Therefore, the cricketer-B is a more consistent.