In statistics, the standard deviation is widely used to find the distance of the mean from the data values. It is frequently used in hypothesis testing and other branches of statistics. The correlation coefficient also used this kind of statistics.

Standard deviation is of two types one is the sample standard deviation and the other is the population standard deviation. Both types are similar in the kinds of variance. In this post, we will learn how to solve problems of standard deviation with examples.

## What is the standard deviation?

The measure of the substance of dispersion or variation of a set of data values is known as the standard deviation. A low result of the standard deviation means that all the values are closer to the expected value (mean or average).

While if the result of the standard deviation is greater then it indicates that the values are far away from the mean. In simple words, the standard deviation measures the closeness or farness of the data values from the expected value.

The set of values can be a sample or population set of data values. The sample set indicates that the set contains some of the observations from the whole population data. While the population indicates that the set contains all the actual data.

## Kinds of standard deviation

There are two kinds of standard deviation.

- Sample standard deviation
- Population standard deviation

Below is a brief introduction to both kinds of standard deviation.

### Sample standard deviation

In statistics, the term population standard deviation is used to find the measure of the dispersion of some of the data from the whole set of data values. In this kind of standard deviation, the values can be calculated by dividing the algebraic sum of squares by the total number of observations minus one.

You have to take the square root of the divided terms to get the result in standard deviation. In other words, the square root of the sample variance is said to be the sample standard deviation.

The general formula for this kind of standard deviation is:

**Sample standard deviation = s = √ [∑ (x _{i} – x̅)^{2}/N – 1]**

**s**= Notation for sample standard deviation.**N**= Total number of population data.**X**= Data values_{i}**x̅**= sample mean**∑ (x**= Statistical sum of squares._{i}– x̅)^{2}

### Population standard deviation

In statistics, the term population standard deviation is used to find the measure of the dispersion of the whole set of data values. In this kind of standard deviation, the values can be calculated by dividing the algebraic sum of squares by the total number of observations.

You have to take the square root of the divided terms to get the result in standard deviation. In other words, the square root of the population variance is said to be the population standard deviation.

The general formula for this kind of standard deviation is:

**Population standard deviation = σ = √ [∑ (x _{i} – μ)^{2}/N]**

**σ**= Notation for population standard deviation.**N**= Total number of population data.**X**= Data values_{i}**μ**= population mean**∑ (x**= Statistical sum of squares._{i}– μ)^{2}

## How to solve problems of standard deviation?

The problems of the standard deviation can be solved easily either by using the formulas of the sample and population or a standard deviation calculator. Here are some examples to learn how to calculate the standard deviation.

### For sample standard deviation

**Example I: **

Find the sample standard deviation of 10, 15, 18, 21, 24, 28, 31

**Solution **

**Step I:** First of all, find the sample mean (x̅).

Sum of sample values = 10 + 15 + 18 + 21 + 24 + 28 + 31

Sum of sample values = 147

Total number of observation = n = 7

Sample mean of data set = x̅ = 147/7

Sample mean of data set = x̅ = 21

**Step II:** Calculate the difference of each data value from the mean.

x_{1} – x̅ = 10 – 21 = -11

x_{2} – x̅ = 15 – 21 = -6

x_{3} – x̅ = 18 – 21 = -3

x_{4} – x̅ = 21 – 21 = 0

x_{5} – x̅ = 24 – 21 = 3

x_{6} – x̅ = 28 – 21 = 7

x_{7} – x̅ = 31 – 21 = 10

**Step III:** Now take the square of deviations to make all the entries of the deviation positive.

(x_{1} – x̅)^{2} = (-11)^{2} = 121

(x_{2} – x̅)^{2} = (-6)^{2} = 36

(x_{3} – x̅)^{2} = (-3)^{2} = 9

(x_{4} – x̅)^{2} = (0)^{2} = 0

(x_{5} – x̅)^{2} = (3)^{2} = 9

(x_{6} – x̅)^{2} = (7)^{2} = 49

(x_{7} – x̅)^{2} = (10)^{2} = 100

**Step IV:** Now find the sum of all the squared deviations.

∑ (x_{i} – x̅)^{2} = 121 + 36 + 9 + 0 + 9 + 49 + 100

∑ (x_{i} – x̅)^{2} = 324

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

∑ (x_{i} – x̅)^{2} / N – 1 = 324 / 7 – 1

∑ (x_{i} – x̅)^{2} / N – 1 = 324 / 6

∑ (x_{i} – x̅)^{2} / N – 1 = 54

**Step VI:** Take the square root of the above expression.

√ [∑ (x_{i} – x̅)^{2} / N – 1] = √54

√ [∑ (x_{i} – x̅)^{2} / N – 1] = 7.348

### For population standard deviation

**Example**

Find the population standard deviation of 7, 9, 10, 12, 18, 22, 26, 32

**Solution **

**Step I:** First of all, find the population mean (μ).

Sum of population values = 7 + 9 + 10 + 12 + 18 + 22 + 26 + 32

Sum of population values = 136

Total number of observation = n = 8

Mean of population data set = μ = 136/8 = 68/4

Mean of population data set = μ = 17

**Step II:** Calculate the difference of each data value from the mean.

x_{1} – μ = 7 – 17 = -10

x_{2} – μ = 9 – 17 = -8

x_{3} – μ = 10 – 17 = -7

x_{4} – μ = 12 – 17 = -5

x_{5} – μ = 18 – 17 = 1

x_{6} – μ = 22 – 17 = 5

x_{7} – μ = 26 – 17 = 9

x_{8} – μ = 32 – 17 = 15

**Step III:** Now take the square of deviations to make all the entries of the deviation positive.

(x_{1} – μ)^{2} = (-10)^{2} = 100

(x_{2} – μ)^{2} = (-8)^{2} = 64

(x_{3} – μ)^{2} = (-7)^{2} = 49

(x_{4} – μ)^{2} = (-5)^{2} = 25

(x_{5} – μ)^{2} = (1)^{2} = 1

(x_{6} – μ)^{2} = (5)^{2} = 25

(x_{7} – μ)^{2} = (9)^{2} = 81

(x_{8} – μ)^{2} = (15)^{2} = 225

**Step IV:** Now find the sum of all the squared deviations.

∑ (x_{i} – μ)^{2} = 100 + 64 + 49 + 25 + 1 + 25 + 81 + 225

∑ (x_{i} – μ)^{2} = 570

**Step V:** Now divide the statistical sum of squares by N.

∑ (x_{i} – μ)^{2} / N = 570 / 8

∑ (x_{i} – μ)^{2} / N = 285 / 4

∑ (x_{i} – μ)^{2} / N = 71.25

**Step VI:** Take the square root of the above expression.

√ [∑ (x_{i} – μ)^{2} / N] = √71.25

√ [∑ (x_{i} – μ)^{2} / N] = 8.441

# Conclusion

In this post, we have covered the basics of the standard deviation and how to calculate it with solved examples. Now you can solve any problem of the standard deviation easily just by grabbing the basics of the above post.