# Mean Deviation and its Coefficient

The mean deviation or the average deviation is defined as the mean of the absolute deviations of observations from some suitable average which may be the arithmetic mean, the median or the mode. The difference $\left( {X - {\text{average}}} \right)$ is called deviation and when we ignore the negative sign, this deviation is written as $\left| {X - {\text{average}}} \right|$ and is read as mod deviations. The mean of these mod or absolute deviations is called the mean deviation or the mean absolute deviation. Thus for sample data in which the suitable average is the$\overline X$, the mean deviation $M.D$ is given by the relation:

For frequency distribution, the mean deviation is given by

When the mean deviation is calculated about the median, the formula becomes

The mean deviation about the mode is

For a population data the mean deviation about the population mean $\mu$ is

The mean deviation is a better measure of absolute dispersion than the range and the quartile deviation.

A drawback in the mean deviation is that we use the absolute deviations $\left| {X - {\text{average}}} \right|$ which does not seem logical. The reason for this is that $\sum (X - \overline X )$ is always equal to zero. Even if we use median or mode in place of $\overline X$, even then the summation $\sum (X - {\text{median}})$ or $\sum (X - {\text{mode}})$will be zero or approximately zero with the result that the mean deviation would always be either zero or close to zero. Thus the very definition of the mean deviation is possible only on the absolute deviations.

The mean deviation is based on all the observations, a property which is not possessed by the range and the quartile deviation. The formula of the mean deviation gives a mathematical impression that is a better way of measuring the variation in the data. Any suitable average among the mean, median or mode can be used in its calculation but the value of the mean deviation is minimum if the deviations are taken from the median. A series drawback of the mean deviation is that it cannot be used in statistical inference.

Coefficient of the Mean Deviation:

A relative measure of dispersion based on the mean deviation is called the coefficient of the mean deviation or the coefficient of dispersion. It is defined as the ratio of the mean deviation to the average used in the calculation of the mean deviation. Thus

Example:

Calculate the mean deviation form (1) arithmetic mean (2) median (3) mode in respect of the marks obtained by nine students gives below and show that the mean deviation from median is minimum.

Marks (out of 25): 7, 4, 10, 9, 15, 12, 7, 9, 7

Solution:

After arranging the observations in ascending order, we get
Marks: 4, 7, 7, 7, 9, 9, 10, 12, 15
${\text{Mean}} = \frac{{\sum X}}{n} = \frac{{80}}{9} = 8.89$
${\text{Median = Value of}}\left( {\frac{{{\text{n + 1}}}}{{\text{2}}}} \right)th{\text{ item = Value of}}\left( {\frac{{{\text{9 + 1}}}}{{\text{2}}}} \right)th{\text{ item}}$
$= {\text{Value of (5)}}th{\text{ item = 9}}$
${\text{ModE = }}7$(Since $7$ is repeated maximum number of times)

 Marks ($X$) $\left| {X - \overline X } \right|$ $\left| {X - {\text{Median}}} \right|$ $\left| {X - {\text{Mode}}} \right|$ $4$ $4.89$ $5$ $3$ $7$ $1.89$ $2$ $0$ $7$ $1.89$ $2$ $0$ $7$ $1.89$ $2$ $0$ $9$ $0.11$ $0$ $2$ $9$ $0.11$ $0$ $2$ $10$ $1.11$ $1$ $3$ $12$ $3.11$ $3$ $5$ $15$ $6.11$ $6$ $8$ Total $21.11$ $21$ $23$

$M.D{\text{ from mean }} = \frac{{\sum \left| {X - \overline X } \right|}}{n} = \frac{{21.11}}{9} = 2.35$

$M.D{\text{ from median }} = \frac{{\sum \left| {X - {\text{Median}}} \right|}}{n} = \frac{{21}}{9} = 2.33$

$M.D{\text{ from mode }} = \frac{{\sum \left| {X - {\text{Mode}}} \right|}}{n} = \frac{{23}}{9} = 2.56$

From the above calculations, it is clear that the mean deviation from the median hast the least value.

Example:

Calculate the mean deviation from mean and its coefficients from the following data.

 Size of Items $3 - 4$ $4 - 5$ $5 - 6$ $6 - 7$ $7 - 8$ $8 - 9$ $9 - 10$ Frequency $3$ $7$ $22$ $60$ $85$ $32$ $8$

Solution:

The necessary calculation is given below:

 Size of Items $X$ $f$ $fX$ $\left| {X - \overline X } \right|$ $f\left| {X - \overline X } \right|$ $3 - 4$ $3.5$ $3$ $10.5$ $3.59$ $10.77$ $4 - 5$ $4.5$ $7$ $31.5$ $2.59$ $18.13$ $5 - 6$ $5.5$ $22$ $121.0$ $1.59$ $34.98$ $6 - 7$ $6.5$ $60$ $390.0$ $0.59$ $35.40$ $7 - 8$ $7.5$ $85$ $637.5$ $0.41$ $34.85$ $8 - 9$ $8.5$ $32$ $272.0$ $1.41$ $45.12$ $9 - 10$ $9.5$ $8$ $76.0$ $2.41$ $19.28$ Total $217$ $1538.5$ $198.53$

${\text{Mean = }}\overline X = \frac{{\sum fX}}{{\sum f}} = \frac{{1538.5}}{{217}} = 7.09$

$M.D{\text{ from mean }} = \frac{{\sum \left| {X - \overline X } \right|}}{n} = \frac{{198.53}}{{217}} = 0.915$

${\text{Coefficient of }}M.D{\text{ (mean) = }}\frac{{M.D{\text{ from Mean}}}}{{{\text{Mean}}}} = \frac{{0.915}}{{7.09}} = 0.129$