Variance and standard deviations are measures of dispersion. In statistics, variance is the average of the squared differences from the mean. Variance is the measurement of deflection of values from its mean. It is denoted by square of a Greek letter sigma ($\sigma ^{2}$). Variance is one of the moments of a distribution and is always non-negative as the squares are positive.
Standard deviation is the root of the sum of the squares of the deviations divided by their number. Also known as root mean square deviation. It is a second moment of a dispersion. Since the sum of the squares of the deviations from the mean is minimum, the deviations are taken only from mean.It is by far the most important and widely used measure of dispersion. It is rigidly defined and based on all the observations. The squaring of the deviations (x - $\bar{x}$) removes the drawback of ignoring the signs of deviations in computing the mean deviation. This step renders it suitable for further mathematical treatment. Moreover, of all the measures of dispersion, standard deviation is affected least by fluctuations of sampling.

Standard deviation gives greater weight to extreme values and as such has not found favor with economists or businessmen who are more interested in the results of the modal class. Standard deviation is considered as the best and the most powerful measure of dispersion.

## Finding Variance and Standard Deviation

The Steps involved in calculating variance and standard deviation are as follows:

• Find the mean for the given data.
• To find the deviation for both periods, for each period find the difference between the average value and the actual value.
• Square each period's deviation and sum the squared deviations.
• Now to find variance divide the sum of squared deviations by the quantity of periods.
• Find the square root of the variance to get the standard deviation.

## Formula for Variance and Standard Deviation

The formula to find variance is given by

$\sigma ^{2}$ = $\frac{\sum (x-\bar{x})^{2}}{n-1}$
Where,
$x$ :  Each Value

$\bar{x}$ : Mean

n : Number of values

$\sum$ : Sum across the values

Formula to find standard deviation is

Individual Observations

$\sigma$ = $\sqrt{\frac{\sum (x-\bar{x})^{2}}{n}}$

Discrete and continuous series

$\sigma$ = $\sqrt{\frac{\sum f(x-\bar{x})^{2}}{n}}$

## Variance and Standard Deviation Examples

### Solved Examples

Question 1: Price of seven different pastries are given below :
24, 30, 45, 56, 75, 100, 65. Find the variance  for the given data.
Solution:

To find the variance we use the formula,

$\bar{x}$ = 56.43, n = 7

$\sigma ^{2}$ = $\frac{\sum (x-\bar{x})^{2}}{n-1}$

= 699.6190

$\sigma$ = 26.450

Therefore, the variance for the given data is 26.450

Question 2: 10 students of B.A. class have obtained the following marks in Maths major out of 100. Calculate the standard deviation of marks obtained.

 Sl.No 1 2 3 4 5 6 7 8 9 10 Marks 5 10 20 25 40 42 45 48 70 80

Solution:

Computation of Standard deviation

 x (x - A) = d d$^{2}$ 5 -35 1225 10 -30 900 20 -20 400 25 -15 225 40 = A 0 0 42 2 4 45 5 25 48 8 64 70 30 900 80 40 1600 $\sum$ = 385 $\sum$d = -15 $\sum$d$^{2}$ = 5343

Total number of terms (n) = 10

$\sigma$ = $\sqrt{\frac{\sum d^{2}}{n}-(\frac{\sum d}{n})^{2}}$

= $\sqrt{\frac{5343}{10}-(\frac{-15}{10})^{2}}$

= $\sqrt{534.3-(-1.5)^{2}}$

= $\sqrt{532.05}$

= 23.066

Therefore, the standard deviation of marks obtained is 23.066.