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:
\[M.D = \frac{{\sum \left {X – \overline X } \right}}{n}\]
For frequency distribution, the mean deviation is given by:
\[M.D = \frac{{\sum f\left {X – \overline X } \right}}{{\sum f}}\]
When the mean deviation is calculated about the median, the formula becomes:
\[M.D{\text{ }}({\text{about median}}) = \frac{{\sum f\left {X – {\text{Median}}} \right}}{{\sum f}}\]
The mean deviation about the mode is:
\[M.D{\text{ }}({\text{about mode}}) = \frac{{\sum f\left {X – {\text{Mode}}} \right}}{{\sum f}}\]
For population data the mean deviation about the population mean $$\mu $$ is:
\[M.D = \frac{{\sum f\left {X – \mu } \right}}{{\sum f}}\]
The mean deviation is a better measure of the absolute dispersion than the range and the quartile deviation.
A drawback of 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 $$\overline X $$, the summation $$\sum (X – {\text{median}})$$ or $$\sum (X – {\text{mode}})$$ will be zero or approximately zero, with the result that the mean deviation will always be either zero or close to zero. Thus the very definition of the mean deviation is possible only for 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 variations 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 the minimum if the deviations are taken from the median. A serious 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 of the average used in the calculation of the mean deviation. Thus:
\[{\text{Coefficient of }}M.D{\text{ (about mean) = }}\frac{{{\text{Mean Deviation from Mean}}}}{{{\text{Mean}}}}\]
\[{\text{Coefficient of }}M.D{\text{ (about median) = }}\frac{{{\text{Mean Deviation from Median}}}}{{{\text{Median}}}}\]
\[{\text{Coefficient of }}M.D{\text{ (about mode) = }}\frac{{{\text{Mean Deviation from Mode}}}}{{{\text{Mode}}}}\]
Example:
Calculate the mean deviation from the (1) arithmetic mean (2) median (3) mode in respect to the marks obtained by nine students given below and show that the mean deviation from the median is the 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 has the least value.
Example:
Calculate the mean deviation from the 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 calculations are 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

$$217$$

$$1538.5$$

$$198.53$$

$${\text{Mean = }}\overline X = \frac{{\sum fX}}{{\sum f}} = \frac{{1538.5}}{{217}} = 7.09$$
$$M.D{\text{ from mean }} = \frac{{\sum \left {X – \overline X } \right}}{n} = \frac{{198.53}}{{217}} = 0.915$$
$${\text{Coefficient of }}M.D{\text{ (mean) = }}\frac{{M.D{\text{ from Mean}}}}{{{\text{Mean}}}} = \frac{{0.915}}{{7.09}} = 0.129$$