In statistics, variance is used for probability distributions. A measure describing how far each number in the set is from the mean. |

Variance basically measures the range in which most of the numbers of a data set are falling. A numerical value indicating how widely individuals in a group are said to vary and is expressed in squared units. It is computed as the average squared deviation of each number from its mean. Variance is denoted by a Greek letter sigma ($\sigma^{2}$). The formulas for variance are given below.

Variance formula for population data is:

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

n = Number of observations

$\bar{x}$ = Population mean

x = Values of observations

Formula for Sample Variance is:

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

n = Number of observations

$\bar{x}$ = Population mean

x = Values of observations

To calculate variance just follows the simple steps given below

From a given data set find its mean:

To find deviation subtract the average value from the actual value.

Square each periods deviation and then sum them.

Variance is now the fraction of the sum of squared deviations by the quantity of periods.

To find the standard deviation just take the square root of the variance.

Variance should have the following properties:

**1)**Variance is said to be non negative.

**2)**Zero variance indicates that all the values are equal in the distribution.

**3)**In a frequency distribution if the values are added by a constant number the variance of the distribution does not change.

Var(Y) = Var(Y + m)

**4)**If the value of a variable is multiplied by a constant then the variance of distribution is multiplied by the square of that constant.

Var(mY) = m$^{2}$ Var(Y). Coefficient of variation is a measure to find the dispersion of data points in a data series around mean. It is defined as the standard deviation $\sigma$ to the mean $\mu$. Even when the means are drastically different from one another we can still use coefficient of variation instead of standard deviation, as it helps in comparing the degree of variation from one data series to another.

It is said to be sensitive to small changes in the mean even if the values do not originate from a ratio scale. Absolute value of the coefficient of variation is said to be relative standard deviation and is expressed in percentages.

**Distributions having coefficient of variation less than 1 are of low variance type and those with coefficient of variation greater than one are of high variance type**.

Equivalent measure to variance is the standard deviation. As compared to the data standard deviation is said to have same dimension. An average of the squared differences from the mean. Standard deviation is denoted by a symbol $\sigma$.

In a normal distribution, about 68% of the scores are said to be in within one standard deviation of the mean.

98% of the scores are within two standard deviation. Many formulas in inferential statistics use the standard deviation.

**Some of the examples based on variance are given below.**

**Example 1:**Weights of few objects were found to be 35kg, 55kg, 25kg, 36kg and 96kg.Find the variance and the standard deviation.

**Solution:**To find mean

Mean = $\frac{35 + 55 + 25 + 36 + 96}{5}$

= $\frac{247}{ 5}$

= 49.4

To find variance find the difference and then square and average the obtained result.

Variance = $\frac{(14.4)^{2}+ (5.6)^{2} + (24.4)^{2} + (13.4)^{2} + 46.6^{2}}{5}$

= $\frac{207.36 + 31.36 + 595.36 + 179.56 + 2171.56}{5}$

= $\frac{3185.2}{5}$

= 637.04

Taking the square root of the above result will give us the standard deviation.

= 25.24

**Example 2:**Calculate the variance of the numbers 5, 3, 9, 6, 7?

**Solution:**

__Step 1:__Mean = $\frac{5+3+9+6+7}{5}$

= $\frac{30}{5}$

= 6

__Step 2:__Subtract the mean value from numbers given in the problem.

=> (5 - 6), (3 - 6), (9 - 6), (6 - 6), (7 - 6)

=> -1, -3, 3, 0, 1

__Step 3:__=> (-1)$^{2}$, (-3)$^{2}$, (3)$^{2}$, 0, 1$^{2}$

=> 1,9, 9, 1, 0

__Step 4:__Variance = $\frac{ (1 + 9 + 9+1 + 0)}{5}$ = $\frac{20}{5}$ = 4

Therefore the variance of a given data set is 4.

**Example 3:**For a group of 200 candidates, the mean and standard deviation of scores were found to be 40 and 15 respectively. Later on it was discovered that the scores 43 and 35 were misread as 34 and 53 respectively. Find the corrected mean and corrected variance corresponding to the corrected figures.

**Solution:**Let x be the given variable. Given n = 200 and $\sigma$ = 15

Now $\bar{x}$ = $\frac{1}{n}$ $\sum_{i = 1}^{n}x_{i}$

$\sum x_{i} = n \bar{x}$

= 200 * 40

= 8000

$\sigma^{2}$= $\frac{1}{n}$$\sum_{i}x_{i}^{2}-\bar{x}^{2}$

or $\sum _{i}x_{i}^{2}=n(\sigma ^{2} + \bar{x}^{2})$

= 200(225 + 1600)

= 365000

Now $\sum x_{i} $ - misread values(mean and variance) + correct values(mean and variance) = 8000 -34 - 53 + 43 +35

= 7991

Corrected Mean = $\frac{7991}{200}$

=39.995

Corrected $\sum _{i}x_{i}^{2} = 365000 -(34)^{2}-(53)^{2}+(43)^{2}+(35)^{2}$

=364109

Corrected $\sigma ^{2}$ = $\frac{364109}{200}$ - $(39.995)^{2}$

= 1820.54 - 1596.40

=224.14