# Standard Deviation

Variance and Standard deviation are the two important topics in Statistics. It is the measure of the dispersion of statistical data. Dispersion is the extent to which values in a distribution differ from the average of the distribution. To quantify the extent of the variation, there are certain measures namely:

(i) Range

(ii) Quartile Deviation

(iii) Mean Deviation

(iv) Standard Deviation

Â The degree of dispersion is calculated by the procedure of measuring the variation of data points. In this article, you will learn what is variance and standard deviation, formulas, and the procedure to find the values with examples.

## What are the Variance and Standard Deviation?

In statistics, Variance and standard deviation are related with each other since the square root of variance is considered the standard deviation for the given data set. Below are the definitions of variance and standard deviation.

### What is variance?

Variance is the measure of how notably a collection of data is spread out. If all the data values are identical, then it indicates the variance is zero.Â  All non-zero variances are considered to be positive. A little variance represents that the data points are close to the mean, and to each other, whereas if the data points are highly spread out from the mean and from one another indicates the high variance. In short, the variance is defined as the average of the squared distance from each point to the mean.

### What is Standard deviation?

Standard Deviation is a measure which shows how much variation (such as spread, dispersion, spread,) from the mean exists. The standard deviation indicates a “typical” deviation from the mean. It is a popular measure of variability because it returns to the original units of measure of the data set.Â  Like the variance, if the data points are close to the mean, there is a small variation whereas the data points are highly spread out from the mean, then it has a high variance. Standard deviation calculates the extent to which the values differ from the average. Standard Deviation, the most widely used measure of dispersion, is based on all values. Therefore a change in even one value affects the value of standard deviation. It is independent of origin but not of scale. It is also useful in certain advanced statistical problems.

## Variance and Standard Deviation Formula

The formulas for the variance and the standard deviation is given below:

Standard Deviation Formula

The population standard deviation formula is given as:

$$\begin{array}{l}\sigma =\sqrt{\frac{1}{N}\sum_{i=1}^{N}(X_i-\mu)^2}\end{array}$$

Here,

Ïƒ = Population standard deviation

N = Number of observations in population

Xi = ith observation in the population

Î¼ = Population mean

Similarly, the sample standard deviation formula is:

$$\begin{array}{l}s =\sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\overline{x})^2}\end{array}$$

Here,

s = Sample standard deviation

n = Number of observations in sample

xi = ith observation in the sample

$$\begin{array}{l}\overline{x}\end{array}$$
= Sample mean

Variance Formula:

The population variance formula is given by:

$$\begin{array}{l}\sigma^2 =\frac{1}{N}\sum_{i=1}^{N}(X_i-\mu)^2\end{array}$$

The sample variance formula is given by:

$$\begin{array}{l}s^2 =\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\overline{x})^2\end{array}$$

## How is Standard Deviation calculated?

The formula for standard deviation makes use of three variables. The first variable is the value of each point within a data set, with a sum-number indicating each additional variable (x, x1, x2, x3, etc). The mean is applied to the values of the variable M and the number of data that is assigned to the variable n. Variance is the average of the values of squared differences from the arithmeticÂ mean.

To calculate the mean value, the values of the data elements have to be added together and the total is divided by the number of data entities that were involved.

Standard deviation, denoted by the symbol Ïƒ,Â describes the square root of the mean of the squares of all the values of a series derived from the arithmetic mean which is also called the root-mean-square deviation. 0 is the smallest value of standard deviation since it cannot be negative. When the elements in a series are more isolated from the mean, then the standard deviation is also large.

The statistical tool of standard deviation is the measures of dispersion that computes the erraticism of the dispersion among the data. For instance, mean, median and mode are the measures of central tendency. Therefore, these are considered to be the central first order averages. The measures of dispersion that are mentioned directly over are averages of deviations that result from the average values, therefore these are called second-order averages.

### Standard Deviation Example

Let’s calculate the standard deviation for the number of gold coins on a ship run by pirates.

There are a total of 100 pirates on the ship. Statistically, it means that the population is 100. We use the standard deviation equation for the entire population if we know a number of gold coins every pirate has.

Statistically, let’s consider a sample of 5 and here you can use the standard deviation equation for this sample population.

This means we have a sample size of 5 and in this case, we use the standard deviation equation for the sample of a population.

Consider the number of gold coins 5 pirates have;Â 4, 2, 5, 8, 6.

Mean:

$$\begin{array}{l}\bar{x} = \frac{\sum x}{n}\end{array}$$

$$\begin{array}{l}=\frac{x_1+x_2+x_3+x_4…..+x_n}{n}\end{array}$$

= (4 + 2 + 5 + 6 + 8) / 5

= 5

$$\begin{array}{l}x_n -\bar{x} \ for\ every\ value\ of\ the\ sample:\end{array}$$

$$\begin{array}{l}x_1 -\bar{x} = 4 – 5 = -1\end{array}$$

$$\begin{array}{l}x_2 -\bar{x} = 2 – 5 = -3\end{array}$$

$$\begin{array}{l}x_3 -\bar{x} = 5 – 5 = 0\end{array}$$

$$\begin{array}{l}x_4 -\bar{x} = 8 – 5 = 3\end{array}$$

$$\begin{array}{l}x_5 -\bar{x} = 6 – 5 = 1\end{array}$$

$$\begin{array}{l}\sum \left ( x_n-\bar{x} \right )^2\end{array}$$

$$\begin{array}{l}= (x_1 -\bar{x})^{2} + (x_2 -\bar{x})^{2}+ â€¦ +(x_5 -\bar{x})^{2}\end{array}$$

$$\begin{array}{l}= (-1)^2 + (-3)^2 + 0^2 + 3^2 + 1^2\end{array}$$

= 20

Standard deviation:

Â

$$\begin{array}{l}S.D = \sqrt{\frac{\sum (x_n-\bar{x})^2}{n-1}}\end{array}$$

$$\begin{array}{l}=\sqrt{\frac{20}{4}}\end{array}$$

=Â âˆš5

= 2.236

### Standard deviation of Grouped Data

In case of grouped data or grouped frequency distribution, the standard deviation can be found by considering the frequency of data values. This can be understood with the help of an example.

Question: Calculate the mean, variance and standard deviation for the following data:

 Class Interval 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 27 10 7 5 4 2

Solution:

 Class Interval Frequency (f) Mid Value (xi) fxi fxi2 0 – 10 27 5 135 675 10 – 20 10 15 150 2250 20 – 30 7 25 175 4375 30 – 40 5 35 175 6125 40 – 50 4 45 180 8100 50 – 60 2 55 110 6050 âˆ‘f = 55 âˆ‘fxi = 925 âˆ‘fxi2 = 27575

N = âˆ‘f = 55

Mean = (âˆ‘fxi)/N = 925/55 = 16.818

Variance = 1/(N – 1) [âˆ‘fxi2 – 1/N(âˆ‘fxi)2]

= 1/(55 – 1) [27575 – (1/55) (925)2]

= (1/54) [27575 – 15556.8182]

= 222.559

Standard deviation = âˆšvariance = âˆš222.559 = 14.918

### Practice Problems on Standard Deviation

1. Calculate the standard deviation of the following values:

5, 10, 25, 30, 50.

2. Find the mean and standard deviation for the following data.

 x 60 61 62 63 64 65 66 67 68 f 2 1 12 29 25 12 10 4 5

3. The diameters of circles (in mm) drawn in a design are given below:

 Diameters 33 – 36 37 – 40 41 – 44 45 – 48 49 – 52 No.of circles 15 17 21 22 25

Calculate the standard deviation and mean diameter of the circles.

[ Hint: First make the data continuous by making the classes as 32.5-36.5, 36.5-40.5,Â  40.5-44.5, 44.5 – 48.5, 48.5 – 52.5 and then proceed.]

Check out more problems on variance and standard deviation of grouped data and Statistics, register with BYJU’S – The Learning App to learn with ease.

## Frequently Asked Questions â€“ FAQs

Q1

### How do you calculate the standard deviation?

The procedure to calculate the standard deviation is given below:
Step 1: Compute the mean for the given data set.
Step 2: Subtract the mean from each observation and calculate the square in each instance.
Step 3: Find the mean of those squared deviations.
Step 4: Finally, take the square root obtained mean to get the standard deviation.
Q2

### What does Standard Deviation tell you?

Standard deviation tells us how far is the mean from each observation in the given data set. In other words, it shows the typical deviation from the mean.
Q3

### What is the standard deviation and variance?

Standard deviation indicates how the spread of observations of a data set is from the mean by studying at the varianceâ€™s square root. The variance estimates the average degree to which each observation differs from the mean of all observations of the data.
Q4

### What is the standard deviation example?

Consider the data set: 2, 1, 3, 2, 4. The mean and the sum of squares of deviations of the observations from the mean will be 2.4 and 5.2, respectively. Thus, the standard deviation will be âˆš(5.2/5) = 1.01.
Q5

### Why do we use standard deviation?

A standard deviation is used to determine how estimations for a group of observations (i.e., data set) are spread out from the mean (average or expected value).