Mean Deviation The Absolute Measure of Dispersion


Statistics / Sunday, September 16th, 2018
(Last Updated On: September 16, 2018)

Mean Deviation Definition

Mean deviation is defined as the mean of absolute deviations of the values from the central value.
For individual series:
\[MD\left( \overline{X} \right)=\frac{\sum{\left| \left( {{X}_{i}}-\overline{X} \right) \right|}}{N}\]
For discrete data with frequency, mean deviation is calculated as:
\[MD\left( \overline{X} \right)=\frac{\sum{f\left| \left( {{X}_{i}}-\overline{X} \right) \right|}}{\sum{f}}\]
In case of continuous series ‘X’ represents mid value of class interval. Similarly, we can have mean deviation from median or mode. X’ is replaced by median or mode in the above formula. However, mean deviation from median is the least. It is known as minimal property of mean deviation.
The corresponding relative measures are coefficient of mean deviation.
\[\text{Coefficient of M}D\left( \overline{X} \right)=\frac{\text{M}D\left( \overline{X} \right)}{\overline{X}}\]

In the following three articles in Statistics we will discuss various types of Absolute Measure of Dispersion in details:
1. Range (R)
2. Quartile Deviation (Q.D.)
3. Standard Deviations (S.D.)

Merits

1. Based on all values.
2. Less affected by extreme values.
3. Not affected by sampling fluctuations

Demerits

1. Not capable of further algebraic treatment.
2. Does not take into account negative signs.

Uses

1. When sampling size is small.
2. In statistical analysis of certain economic, business and social phenomena.

 Example 01

Compute MD and Coefficient of MD from mean for the following data: 21, 32, 38, 41, 49, 54, 59, 66, 68

Solution:
\[Mean\left( \overline{X} \right)=\frac{\sum\limits_{i=1}{{{X}_{i}}}}{N}=\frac{428}{9}=47.55\]

X \[d=\left| {{X}_{i}}-\overline{X} \right|\]
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
\[\sum{X=428}\] \[\sum{\left| {{X}_{i}}-\overline{X} \right|}=\sum{d=116.45}\]

\[MD=\frac{\sum{\left| {{X}_{i}}-\overline{X} \right|}}{N}=\frac{116.45}{9}=12.94\]
\[\text{Coefficient of M}D\left( \overline{X} \right)=\frac{\text{M}D\left( \overline{X} \right)}{\overline{X}}\]
\[\Rightarrow \text{Coefficient of M}D\left( \overline{X} \right)=\frac{12.94}{47.55}=0.272\]

 Example 02

Following are the wages of workers. Find mean deviation from median and its coefficient.

X 59 32 67 43 22 17 64 55 47 80 25
Wages 17 22 25 32 43 47 55 59 64 67 80

Solution:

\[Median={{\left( \frac{11+1}{2} \right)}^{th}}item={{6}^{th}}item=47\]

X Wages (f) \[\left| {{X}_{i}}-Me \right|=\left| {{X}_{i}}-47 \right|\] \[{{f}_{i}}.\left| {{X}_{i}}-Me \right|\]
59 17 30 510
32 22 25 550
67 25 22 550
43 32 15 480
22 43 4 172
17 47 ← Me 0 0
64 55 8 440
55 59 12 708
47 64 17 1088
80 67 20 1340
25 80 33 2640
\[\sum{{{f}_{i}}}=511\] \[\sum{\left| {{X}_{i}}-Me \right|=186}\] \[\sum{{{f}_{i}}.\left| {{X}_{i}}-Me \right|}=8478\]

\[MD=\frac{\sum{{{f}_{i}}.\left| {{X}_{i}}-Me \right|}}{\sum{{{f}_{i}}}}=\frac{8478}{511}=16.591\]
\[\text{Coefficient of M}D=\frac{MD}{Median}\]
\[\Rightarrow \text{Coefficient of M}D=\frac{16.591}{47}=0.353\]

 Example 03

Compute mean deviation about its mode and its coefficient.

X 20 40 60 80 100 120 140 160 180
f 6 19 40 23 65 83 55 20 9

Solution:
The highest frequency is 83 and hence mode is 120.

X f \[d=\left| {{X}_{i}}-Mode \right|\] fd
20 6 100 600
40 19 80 1520
60 40 60 2400
80 23 40 920
100 65 20 1300
120 83 0 0
140 55 20 1100
160 20 40 800
180 9 60 540
\[N=\sum{{{f}_{i}}=320}\] \[\sum{{{f}_{i}}.\left| {{X}_{i}}-Mode \right|}=9180\]

\[MD=\frac{\sum{{{f}_{i}}.\left| {{X}_{i}}-Mode \right|}}{N}=\frac{9180}{320}=28.69\]
\[\text{Coefficient of M}D=\frac{MD}{Mode}=\frac{28.69}{120}=0.239\]

 Example 04

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

Solution:
Here N = 22+18+10+8+2 = 60
\[Median={{\left( \frac{60+1}{2} \right)}^{th}}item={{30.5}^{th}}item\]
It lies in 40 cumulative frequency (cf) and against 40 cf discrete value is 50.

X No. of Employees x( Mid value) Cumulative frequency (cf) \[d=\left| {{x}_{i}}-Me \right|\] 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
\[N=\sum{{{f}_{i}}}=60\] \[\sum{{{f}_{i}}.\left| {{x}_{i}}-Me \right|}=1770\]

\[MD=\frac{\sum{{{f}_{i}}.\left| {{X}_{i}}-Me \right|}}{\sum{{{f}_{i}}}}=\frac{1770}{60}=29.5\]
\[\text{Coefficient of M}D=\frac{MD}{Median}=\frac{29.5}{50}=0.59\]

Leave a Reply

Your email address will not be published. Required fields are marked *