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 β Lowest value
10. Let us consider a set of observations π₯1, π₯2,π₯3, β¦β¦β¦β¦β¦β¦ , π₯ π and π π» is
maximum and π πΏ is minimum.
Then Range = π π» β π πΏ.
Example:
Find out the range of the set of observations, -7, -2, -4, 0, 8.
Solution:
Here, maximum value, π π» = 8 and minimum value, π πΏ = β7
Range = π π» β π πΏ = 8 β β7 = 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 = π π β π πΏ
Where,
π π= The upper boundary of the highest class.
π πΏ= 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, π π = ππΎ. 2200
And the lowest boundary of the highest class, π πΏ = ππΎ. 1700
Then Range = π π β π πΏ = ππΎ. 2200 β ππΎ. 1700 = ππΎ. 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 π₯1, π₯2,π₯3, β¦β¦β¦β¦β¦β¦ , π₯ π 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 (π) =
π₯ π;π₯π
π=1
π
; π = π΄πππ‘ππππ‘ππ ππππ
ii. Mean deviation about median, M.D (π) =
π₯ π;πππ
π=1
π
; ππ = ππππππ
iii.Mean deviation about mode, M.D (π) =
π₯ π;πππ
π=1
π
; ππ = ππππ
18. Example:
Find out the mean deviation from the following given set 2,3,4,5,6.
Solution: We know,
Mean deviation, M.D (π) =
π₯ π;π₯π
π=1
π
Here, π =
π₯ π
π
π=1
π
=
2:3:4:5:6
5
= 4
Mean deviation, M.D (π) =
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) π₯ β π₯ = π₯ β 102
103 π₯ β 102 = 103 β 102 = 1
97 π₯ β 102 = 97 β 102 = 5
101 1
106 4
103 1
π₯π = 510 π₯π β π₯ = 12
Arithmetic Mean,
π₯ =
π₯π
π
π<1
π
=
510
5
= 102
Mean Deviation about mean, M.D (π) =
π₯ π;π₯π
π=1
π
=
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 π₯1, π₯2,π₯3, β¦β¦β¦β¦β¦β¦ , π₯ π occur with frequencies π1, π2,π3, β¦β¦β¦β¦β¦β¦ , ππ
respectively then the mean deviation can be written as:
i. Mean deviation about arithmetic mean, M.D (π) =
π π π₯ π;π₯π
π=1
π
ii. Mean deviation about median, M.D (π) =
π π π₯ π;πππ
π=1
π
iii.Mean deviation about mode, M.D (π) =
π π π₯ π;πππ
π=1
π
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)
ππ π β π π π β π
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 ππ = 50 ππ π₯π = 1550 ππ π₯π β π₯ = 688
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, π =
π π’π ππ π ππ’ππππ ππ πππ£πππ‘πππ ππππ ππππ
πππ‘ππ ππ. ππ πππ πππ£ππ‘πππ
28. 1. Standard Deviation for Ungrouped Data:
Let,π₯1, π₯2,π₯3, β¦β¦β¦β¦β¦β¦ , π₯ π be a set of n observations or values, then their
standard deviation can be written as,
Standard Deviation, ππ₯ =
(π₯ π
π
π=1 ;π₯)2
π
=
π₯ π
2
π
β (
π₯ π
π
)2
29. 2. Standard Deviation for Grouped Data:
If π₯1, π₯2,π₯3, β¦β¦β¦β¦β¦β¦ , π₯ π occur with frequencies π1, π2,π3, β¦β¦β¦β¦β¦β¦ , ππ
respectively then the standard deviation can be written as:
Standard Deviation, ππ₯ =
π π(π₯ π
π
π=1 ;π₯)2
π
=
π π π₯ π
2
π
β (
π π π₯ π
π
)2
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 (π3and lower quartile π1by 2.
Mathematically, the quartile deviation (Q.D) can be written as:
π. π· =
πππππ ππ’πππ‘πππ;πΏππ€ππ ππ’πππ‘πππ
2
=
π3;π1
2
The term (π3 β π1) is known as the interquartile range and the quartile deviation
π3;π1
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, π1 =
π
4
π‘π πππ πππ£ππ‘πππ:(
π
4
:1)π‘π πππ πππ£ππ‘πππ
2
π1 =
8
4
π‘π πππ πππ£ππ‘πππ:(
8
4
:1)π‘π πππ πππ£ππ‘πππ
2
π1 =
2ππ πππ πππ£ππ‘πππ:3ππ πππ πππ£ππ‘πππ
2
π1 =
30:40
2
= 35
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,
π1 =
π
4
π‘π πππ πππ£ππ‘πππ
=
29
4
π‘π πππ πππ£ππ‘πππ
= 7.25 π‘π πππ πππ£ππ‘πππ
Therefore, 7.25 th observation is first quartile
which is lies in the class 10-15.
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
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:
πΆ. π =
π
π₯
Γ 100; π₯ β 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
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) π π Score (y) π π
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
π₯ = 580 π₯2
= 71328 π¦ = 200 π¦2
= 4626
58. We know, Coefficient of Variation, πΆ. π =
π
π₯
Γ 100
For Cricketer-A
Arithmetic Mean,
π =
π
π
=
πππ
ππ
= ππ
Standard Deviation,
π =
π π
π
β (
π
π
) π
=
πππππ
ππ
β (
πππ
ππ
) π
= ππ. ππ
Coefficient of Variation,
πͺ. π½ =
π
π
Γ πππ =
ππ. ππ
ππ
Γ πππ
= πππ. ππ%
For Cricketer-B
Arithmetic Mean,
π =
π
π
=
πππ
ππ
= ππ
Standard Deviation,
π =
π π
π
β (
π
π
) π
=
ππππ
ππ
β (
πππ
ππ
) π
= π. ππ
Coefficient of Variation,
πͺ. π½ =
π
π
Γ πππ =
π. ππ
ππ
Γ πππ
= ππ. ππ%
Comment:
From the above result, we see that, C.V. (A) = 105.84% and C.V. (B) 39.56%
Since, C.V. (A)> π. π. (π). Therefore, the cricketer-B is a more consistent.