This document discusses measures of dispersion such as mean deviation. It provides formulas to calculate mean deviation from the mean, median, and mode. It also gives examples of calculating mean deviation from given data sets and interpreting the results. Mean deviation is the average of the absolute deviations from the mean, median, or mode. It indicates how far the values are spread out from the central tendency. The document shows how to compute mean deviation and its coefficient from sample data for different measures of central tendency.

1. Session – 8
Measures of Dispersions
Mean Deviation
Mean deviation is the average differences among the items in a series from the
mean itself or median or mode of that series. It is concerned with the extent of which
the values are dispersed about the mean or median or the mode. It is found by
averaging all the deviations from control tendency. These deviations are taken into
computations with regard to negative sign. Theoretically the deviations of item are
taken preferably from median instead than from the mean and mode.
Merits of Mean Deviation
 It is rigidly defined and easy to compute.
 It takes all items in to considerations and gives weight to deviation according to
these sign.
 It is less affected by extreme values.
 It removes all irregularities by obtaining deviation and provides correct measures.
Demerits of Mean Deviation
 It is not suitable for algebraic treatments.
 It is positive which is not justified mathematically.
 It is not satisfactory measure when the deviations are taken from mode.
 It is not suitable when class intervals are open end.

2. Formula to compute Mean Deviation
If xi is variant and takes the values x1, x2, x3, …….. xn with average. A (mean,
median, mode), then mean deviation from the average – A is defined by
MD =
x A i  
N
For the grouped data
MD =
f x A i  
N
Coefficient of MD =
MD
Mean
1. Compute MD and CMD from mean for the given data below.
X d = x x i 
21 26.55
32 15.55
38 9.55
41 6.55
49 1.45
54 6.45
59 11.45
66 18.45
68 20.45
x = 428  x x i  = d= 116.45


428
x  
x n x 47.35
i 1
i
9
MD =
116.45
9
 
x x i 
N
MD =  = 12.938
Coefficient of MD =
MD
Avg
=
12.938
47.55
= 0.272

3. 2. Following are the wages of workers. Find mean deviation from median and its
coefficient.
x Wages x Me i  = x 47 i 
59 17 30
32 22 25
67 25 22
43 32 15
22 43 4
17 47  M 0
64 55 8
55 59 12
47 64 17
80 67 20
25 80 33
25  x M i  = 186  x M i  = 186

 
11 1 th item
Median = 

2

 
11 1
= 

2
= 6th item
Me = 47
MD =
x Me i  
N
=
186
11
= 16.91
Coefficient of MD =
MD
Median
16.91  = 0.359
47

4. 3. Compute MD about its mode and its coefficient.
x f d = x Mode i  fd
20 6 100 600
40 19 80 1520
60 40 60 2400
80 23 40 920
100 65 20 1300
120  Mode 83  Modal
class
0 0
140 55 20 1100
160 20 40 800
180 9 60 5401
f = 320 f x Mode i 
= 9180
the highest frequency is 83 and hence
Z = 120
MD=
x Mode i  
N


320
Median = 

9180
= 28.68
Coefficient of MD =
28.68
120
= 0.239

5. 4. Find out the mean deviation from the data given below about its median.
Salaries 40 50 50-100 100-200 200-400
No. of Employees 22 18 10 8 2
x No. of
Employees
x(mv) cf d = x Me i  fd
40 22 40 22 10 220
50 18 50 40 0 0
50-100 10 75 50 25 250
100-200 8 150 58 100 800
200-400 2 300 60 250 500
f = 60 f x Me i  = 1770
Median =
th
 
N 1
2



item
=
60  1
2
=
61
2
= 30.5 It lies in 40 cf and against 40 cf
discrete value is 50
MD =
x Median i  
N


60
= 

1770
MD = 29.5
Coefficient of MD =
MD
Median
=
29.5
50
= 0.59