# 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 $d=\left| {{X}_{i}}-\overline{X} \right|$ 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 $N=\sum{{{f}_{i}}}=60$ $\sum{{{f}_{i}}.\left| {{x}_{i}}-Me \right|}=1770$

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

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