**Standard deviation and variance :**

Although mean deviation is an improvement over range so far as a measure of dispersion is concerned, mean deviation is difficult to compute and further more, it cannot be treated mathematically.

The best measure of dispersion is, usually, standard- deviation which does not possess the demerits of range and mean deviation.

SD for a given set of observations is defined as the root mean square deviation when the deviations are taken from the AM of the observations.

Let the variable "x" assume "n" values as given below

Then, the formula for SD is given by

For a grouped frequency distribution, the SD is given by

Sometimes the square of SD, known as variance by taking square of standard deviation, is regarded as a measure of dispersion.

We have, then,

For a grouped frequency distribution, the variance is given by

A relative measure of dispersion using SD is given by coefficient of variation (cv) which is defined as the ratio of standard-deviation to the corresponding arithmetic mean, expressed as a percentage.

1) If all the observations assumed by a variable are constant i.e. equal, then the SD is zero. This means that if all the values taken by a variable x is k, say , then s = 0. This result applies to range as well as mean deviation.

2) SD remains unaffected due to a change of origin but is affected in the same ratio due to a change of scale i.e., if there are two variables x and y related as y = a+bx for any two constants a and b, then SD of y is given by

SD of "y" = |b| x SD of "x"

3) If there are two groups containing n₁ and n₂ observations, x̄₁ and x̄₂ as respective arithmetic means, S₁ and S₂ as respective standard deviations, then combined SD is given by

This result can be extended to more than 2 groups.

**Problem 1 :**

Find the SD and the coefficient of variation for the following numbers:

5, 8, 9, 2, 6

**Solution :**

**Computation of SD**

The formula to find SD for the given data is

The coefficient of variation is

CV = 100 x SD / AM

CV = 100 x 2.45 / 6

**CV = 40.83**

**Problem 2 : **

Find the SD of the following distribution:

**Solution :**

The formula to find SD for the given data is

**Problem 3 :**

If AM and coefficient of variation of x are 10 and 40 respectively, what is the variance of (15–2x) ?

**Solution :**

Let y = 15 - 2x

When "x" and "y" are related as y = a + bx, then

SD of "y" = |b| x SD of "x"

SD of "y" = |-2| x SD of "x"

SD of "y" = 2 x SD of "x" -------(1)

Coefficient of variation of "x" = 40 and mean of "x" = 10

Coefficient of variation of "x" = 40

100 x SD / AM = 40

100 x SD / 10 = 40

10 x SD = 40

**SD of "x" = 4**

(1) -------> SD of "y" = 2 x 4

SD of "y" = 8

Variance of "y" = 8²

Variance of "y" = 64

**Hence, variance of "15-2x" is 64**

We may now have a review of the different measures of dispersion on the basis of their relative merits and demerits.

SD, like AM, is the best measure of dispersion. It is rigidly defined, based on all the observations, not too difficult to compute, not much affected by sampling fluctuations and moreover it has some desirable mathematical properties.

All these merits of standard deviation make SD as the most widely and commonly used measure of dispersion.

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