# What is Variance? Its Definition, Kinds, and Solved Examples

The variance is frequently used in statistics to measure the typical distance of all the given observations from the mean. The variance is usually used to measure various kinds of statistics and hypothesis testing.

It is a general technique to find the closeness and farness of the data values from the average value. In this post, we will learn the basics of variance such as the definition, formulas, and solved examples.

**What is the Variance?**

In statistics, the variance is generally used to measure how far a set of observations is spread out from the expected value. It is also defined as the measure of dispersion. The set of observations is of two types; one is the sample data set while the other is the population data set.

The set of population data values is the set of the whole set of observations, while the sample set of data values is the set of some of the observations from the whole.

**Kinds of Variance**

The variance is of two kinds:

- Sample variance
- Population variance

Let’s briefly describe the kinds of variance.

### Population Variance

The population variance is the first type of variance that deals with the whole set of observations. It is denoted by “**σ**^{2}”.

It is usually used to measure the dispersion of the data values from the mean. After finding the square of deviations of population data values, divide it by the total number of observations to measure the output of the population variance.

Following is a general formula of the population variance.

**Population variance = σ**^{2}** = ∑ (x**_{i}** – μ)**^{2}**/N**

** σ**^{2} = Denotes the population variance-
**N **= Total number of observations in a given data set. **x**_{i}** **= Data values. **μ** = population mean- (x
_{i} – μ)^{2} = statistical sum of squares

### Sample Variance

The sample variance is the other type of variance that deals with the set of some of the observations from the whole. It is denoted by “**s**^{2}”.

It is usually used to measure the dispersion of the data values from the mean. After finding the square of deviations of sample data values, divide it by the total number of observations minus one to measure the output of the population variance.

Following is a general formula of the sample variance.

**Sample variance = s**^{2}** = ∑ (x**_{i}** – x̅)**^{2}**/N – 1**

**s**^{2} = Denotes the sample variance-
**N **= Total number of observations in a given data set. **x**_{i}** **= Data values. **x̅** = population mean - (x
_{i} – **x̅**)^{2} = statistical sum of squares

**How to Calculate Problems of the Variance?**

The problems of the variance can be calculated easily by using the formulas of population and sample variances. Following are a few solved examples of the variance to learn how to calculate the problems of the variance.

**For Sample Variance **

**Example I: **

Evaluate the variance of the given sample data,

11, 15, 16, 20, 33, 37, 40, 44, 59, 65, 70, 82

**Solution **

**Step 1:** Take the given set of sample data values and find the sample mean.

Sample data = 11, 15, 16, 20, 33, 37, 40, 44, 59, 65, 70, 82

Sum of sample data = ∑x = 11 + 15 + 16 + 20 + 33 + 37 + 40 + 44 + 59 + 65 + 70 + 82

Sum of sample data = ∑x = 492

Total number of observation = n = 12

Sample mean of data set = x̅ = ∑x/n = 492/12 = 246/6

Sample mean of data set = x̅ = ∑x/n = 41

**Step 2:** Now find the difference between the sample values from the sample mean.

x_{1} – x̅ = 11 – 41 = -30

x_{2} – x̅ = 15 – 41 = -26

x_{3} – x̅ = 16 – 41 = -25

x_{4} – x̅ = 20 – 41 = -21

x_{5} – x̅ = 33 – 41 = -8

x_{6} – x̅ = 37 – 41 = -4

x_{7} – x̅ = 40 – 41 = -1

x_{8} – x̅ = 44 – 41 = 3

x_{9} – x̅ = 59 – 41 = 18

x_{10} – x̅ = 65 – 41 = 24

x_{11} – x̅ = 70 – 41 = 29

x_{12} – x̅ = 82 – 41 = 41

**Step 3:** To make the positive deviations, calculate the square of differences of data values from the expected value.

(x_{1} – x̅)^{2} = (-30)^{2} = 900

(x_{2} – x̅)^{2} = (-26)^{2} = 676

(x_{3} – x̅)^{2} = (-25)^{2} = 625

(x_{4} – x̅)^{2} = (-21)^{2} = 441

(x_{5} – x̅)^{2} = (-8)^{2} = 64

(x_{6} – x̅)^{2} = (-4)^{2} = 16

(x_{7} – x̅)^{2} = (-1)^{2} = 1

(x_{8} – x̅)^{2} = (3)^{2} = 9

(x_{9} – x̅)^{2} = (18)^{2} = 324

(x_{10} – x̅)^{2} = (24)^{2} = 576

(x_{11} – x̅)^{2} = (29)^{2} = 841

(x_{12} – x̅)^{2} = (41)^{2} = 1681

**Step 4:** Now find the sum of squared deviations.

∑ (x_{i} – x̅)^{2} = 900 + 676 + 625 + 441 + 64 + 16 + 1 + 9 + 324 + 576 + 841 + 1681

∑ (x_{i} – x̅)^{2} = 6154

**Step 5:** Now divide the statistical sum of squares by N – 1.

∑ (x_{i} – x̅)^{2} / N – 1 = 6154 / 12 – 1

∑ (x_{i} – x̅)^{2} / N – 1 = 6154 / 11

∑ (x_{i} – x̅)^{2} / N – 1 = 559.46

A variance calculator can be used to find the result of the above problem to ease up the calculations.

**For Population Variance**

**Example II: **

Evaluate the variance of the given population data,

10, 25, 36, 40, 43, 44, 47, 50, 56

**Solution **

**Step 1:** Take the given set of population data values and find the population means.

Population data = 10, 25, 36, 40, 43, 44, 47, 50, 56

Sum of population values = ∑x = 10 + 25 + 36 + 40 + 43 + 44 + 47 + 50 + 56

Sum of population values = ∑x = 351

Total number of observation = n = 9

Mean of population data set = μ = ∑x/n = 351/9 = 117/3

Mean of population data set = μ = ∑x/n = 39

**Step 2:** Now find the difference between the population values from the population mean.

x_{1} – μ = 10 – 39 = -29

x_{2} – μ = 25 – 39 = -14

x_{3} – μ = 36 – 39 = -3

x_{4} – μ = 40 – 39 = 1

x_{5} – μ = 43 – 39 = 4

x_{6} – μ = 44 – 39 = 5

x_{7} – μ = 47 – 39 = 8

x_{8} – μ = 50 – 39 = 11

x_{9} – μ = 56 – 39 = 17

**Step 3:** To make the positive deviations, calculate the square of differences of data values from the expected value.

(x_{1} – μ)^{2} = (-29)^{2} = 841

(x_{2} – μ)^{2} = (-14)^{2} = 196

(x_{3} – μ)^{2} = (-3)^{2} = 9

(x_{4} – μ)^{2} = (1)^{2} = 1

(x_{5} – μ)^{2} = (4)^{2} = 16

(x_{6} – μ)^{2} = (5)^{2} = 25

(x_{7} – μ)^{2} = (8)^{2} = 64

(x_{8} – μ)^{2} = (11)^{2} = 121

(x_{9} – μ)^{2} = (17)^{2} = 289

**Step 4:** Now find the sum of squared deviations.

∑ (x_{i} – μ)^{2} = 841 + 196 + 9 + 1 + 16 + 25 + 64 + 121 + 289

∑ (x_{i} – μ)^{2} = 1562

**Step 5:** Now divide the above summation value by N.

∑ (x_{i} – μ)^{2} / N = 1562 / 9

∑ (x_{i} – μ)^{2} / N = 173.56

You can verify the above result by using a variance calculator.

**Conclusion**

In this article, we have learned how to calculate the problems of the variance along with solved examples. The content of this post must help you in understanding the variance more accurately.