Mean Deviation The Absolute Measure of Dispersion


Statistics / Sunday, September 16th, 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\[d=\left| {{X}_{i}}-\overline{X} \right|\]
2126.55
3215.55
389.55
416.55
491.45
546.45
5911.45
6618.45
6820.45
\[\sum{X=428}\]\[\sum{\left| {{X}_{i}}-\overline{X} \right|}=\sum{d=116.45}\]

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

 Example 02

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

X5932674322176455478025
Wages1722253243475559646780

Solution:

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

XWages (f)\[\left| {{X}_{i}}-Me \right|=\left| {{X}_{i}}-47 \right|\]\[{{f}_{i}}.\left| {{X}_{i}}-Me \right|\]
591730510
322225550
672522550
433215480
22434172
1747 ← Me00
64558440
555912708
4764171088
8067201340
2580332640
\[\sum{{{f}_{i}}}=511\]\[\sum{\left| {{X}_{i}}-Me \right|=186}\]\[\sum{{{f}_{i}}.\left| {{X}_{i}}-Me \right|}=8478\]

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

 Example 03

Compute mean deviation about its mode and its coefficient.

X20406080100120140160180
f6194023658355209

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

Xf\[d=\left| {{X}_{i}}-Mode \right|\]fd
206100600
4019801520
6040602400
802340920
10065201300
1208300
14055201100
1602040800
180960540
\[N=\sum{{{f}_{i}}=320}\]\[\sum{{{f}_{i}}.\left| {{X}_{i}}-Mode \right|}=9180\]

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

 Example 04

Find out the mean deviation from the data given below about its median.

Salaries405050-100100-200200-400
No. of Employees22181082

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.

XNo. of Employeesx( Mid value)Cumulative frequency (cf)\[d=\left| {{x}_{i}}-Me \right|\]fd
4022402210220
5018504000
50-10010755025250
100-200815058100800
200-400230060250500
\[N=\sum{{{f}_{i}}}=60\]\[\sum{{{f}_{i}}.\left| {{x}_{i}}-Me \right|}=1770\]

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

 

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