In statistics, the measure of dispersion is defined as the measurement of the extent to which the statistical data is dispersed or spread. It is the measure of the range of stretch or squeeze of the distribution. Dispersion is also termed as scatter, variability or spread.

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The virtue of a distribution to spread over a range of data is called the dispersion. In statistics, it is important to know the pattern of dispersion in order to study the data. The ways by which the dispersion is computed are known as the measures of dispersion. It basically quantifies the spread or squeeze of the values in a distribution.

There are several different types of measures of dispersion. Most important ones are listed below:

Quartile deviation is a method of measuring dispersion using lower and upper quartiles.

The range is calculated by finding the difference between smallest and largest values of an observation.

Average of the absolute differences of all the values in a data from the mean is known as mean deviation or average derivation.

Standard deviation is also an important measure of dispersion about the mean of the data.

Variance is determined by calculating the square of standard deviation.

Range = $L - S$

Coefficient of range = $\frac{L-S}{L+S}$

Where, $L$ and $S$ are the largest and smallest values.

Quartile Deviation = $\frac{(Q_3 - Q_1)}{ 2}$

Coefficient of Quartile Deviation = $\frac{(Q_3-Q_1)}{ Q_3+Q_1}$

Where, $Q_3$ and $Q_1$ being third and first quartiles.

$MD$ = $\frac{\sum _{i}^{n} |x_i-\mu|}{n}$

$MD$ = $\frac{\sum _{i}^{n} f_i|x_i-\mu|}{n}$

Where, $\mu$ = Mean

$x_i$ = Data Values

$f_i$ = Corresponding Frequencies

SD = $\sqrt{\frac{\sum_{i}^{n}(x_i-\mu)^2}{n-1}}$

Where, $\mu$ = mean

$x_i$ = data values

Variance $\sigma$ = $SD^2$

$64, 75, 59, 89, 92, 96, 67, 58, 82, 71, 60, 90, 63$

Calculate range.

Largest score = $96$

Range = $96 - 58$ = $38$

Find the standard deviation of the following data.

$8, 4, 5, 2, 6$

$\mu$ = $\frac{8+4+5+2+6}{5}$ = $\frac{25}{5}$ = $5$

$SD$ = $\sqrt{\frac{\sum_{i}^{n}(x_i-\mu)^2}{n-1}}$

= $ \sqrt{\frac{(8-5)^2+(4-5)^2+(5-5)^2+(2-5)^2+(6-5)^2}{5-1}}$

= $\sqrt{\frac{(3)^2+(-1)^2+(0)^2+(-3)^2+(1)^2}{4}}$

= $\sqrt{\frac{9+1+0+9+1}{4}}$

= $\sqrt{5}$ = $2.24$