Mean Deviation The Absolute Measure of Dispersion

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 the case of continuous series ‘X’ represents the mid-value of the 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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