Variance and 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

 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.  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.  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:

$\begin{array}{l}�=\sqrt{\frac{1}{�}\sum _{�=1}^{�}({�}_{�}-�{)}^2}\end{array}$

Here,

σ = Population standard deviation

N = Number of observations in population

Xi = ith observation in the population

μ = Population mean

Similarly, the sample standard deviation formula is:

$\begin{array}{l}�=\sqrt{\frac{1}{�-1}\sum _{�=1}^{�}({�}_{�}-\overline{�}{)}^2}\end{array}$

Here,

s = Sample standard deviation

n = Number of observations in sample

xi = ith observation in the sample

$\begin{array}{l}\overline{�}\end{array}$

= Sample mean

Variance Formula:

The population variance formula is given by:

$\begin{array}{l}{�}^2=\frac{1}{�}\sum _{�=1}^{�}({�}_{�}-�{)}^2\end{array}$

The sample variance formula is given by:

$\begin{array}{l}{�}^2=\frac{1}{�-1}\sum _{�=1}^{�}({�}_{�}-\overline{�}{)}^2\end{array}$

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 arithmetic 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 σ, 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; 4, 2, 5, 8, 6.

Mean:

$\begin{array}{l}\overline{�}=\frac{\sum �}{�}\end{array}$

$\begin{array}{l}=\frac{{�}_1+{�}_2+{�}_3+{�}_4\dots ..+{�}_{�}}{�}\end{array}$

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

= 5

$\begin{array}{l}{�}_{�}-\overline{�}����������������ℎ�������:\end{array}$

$\begin{array}{l}{�}_1-\overline{�}=4–5=-1\end{array}$

$\begin{array}{l}{�}_2-\overline{�}=2–5=-3\end{array}$

$\begin{array}{l}{�}_3-\overline{�}=5–5=0\end{array}$

$\begin{array}{l}{�}_4-\overline{�}=8–5=3\end{array}$

$\begin{array}{l}{�}_5-\overline{�}=6–5=1\end{array}$

$\begin{array}{l}\sum {({�}_{�}-\overline{�})}^2\end{array}$

$\begin{array}{l}=({�}_1-\overline{�}{)}^2+({�}_2-\overline{�}{)}^2+\dots +({�}_5-\overline{�}{)}^2\end{array}$

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

= 20

Standard deviation:

 

$\begin{array}{l}�.�=\sqrt{\frac{\sum ({�}_{�}-\overline{�}{)}^2}{�-1}}\end{array}$

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

= √5

= 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
	∑f = 55		∑fxi = 925	∑fxi2 = 27575

N = ∑f = 55

Mean = (∑fxi)/N = 925/55 = 16.818

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

= 1/(55 – 1) [27575 – (1/55) (925)2]

= (1/54) [27575 – 15556.8182]

= 222.559

Standard deviation = √variance = √222.559 = 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,  40.5-44.5, 44.5 – 48.5, 48.5 – 52.5 and then proceed.]

&

Win