Overview
Test Series
In this lesson we will investigate the concepts of variance and standard deviation when working with data in statistics. These two concepts help us with our understanding of how the distribution of values is dispersed from the average. We will cover what variance and standard deviation are, we will introduce stepwise formulas, and cover helpful hints to apply them.Along with this, we will also cover their main properties and answer some common FAQs for better understanding. To make learning more practical, a few solved examples are included so that students can clearly see how variance and standard deviation are calculated and can confidently use them in exams.
The primary difference between std deviation and variance is that standard deviation is expressed in the same units as the mean of data, whereas the variance is expressed in squared units.
Variance and standard deviation in mathematics can be determined by employing the mean of a group of numbers in question. The mean is defined as the average of a collection of numbers, moreover, the variance estimates the average degree to which each number is different from the mean.
Variance: The variance implies the average of the squared differences from the mean. To calculate the variance, first, determine the difference between each position and the mean; then, square and average the outcomes.
In common terms, variance is a measure of how far a set of data/numbers are dispersed out from their mean/ average value. It is denoted by the symbol \(σ^{2}\).
Standard Deviation: Standard deviation is a statistic that studies how notably from the mean a group of numbers is, by applying the square root of the variance.
The estimation of variance uses squares because it measures outliers more heavily than data closer to the mean. This sort of calculation also limits differences above the mean from cancelling out those below, which would end in a variance of zero.
In simple terms, the spread of statistical data is estimated by the standard deviation. Distribution measures the deviation of data/information from its mean/average state. The degree of dispersion is calculated by the method of estimating the deviation of data points. Standard deviation is expressed by the symbol, ‘σ’.

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Variance of the data set is the average square distance between the mean value and each specific data value. Plus standard deviation explains the spread of data values around the mean. With this knowledge let us learn about the standard deviation and variance formula.
When we have gathered data from every portion of the population then we are interested to get an exact value for population variance.
Population variance is given by the formula:
\(σ^{2}=\frac{1}{N}\sum_{i=1}^N(X_i−μ)^{2}\)
Where:
\(σ^2\)=Population variance
N = Number of observations in the population.
\(X_i\)=ith observation in the population.
\(\mu\)=Population mean or Assumed mean.
Check out this article on Covariance.
When we accumulate data from a sample, the sample variance is applied to make estimates or conclusions about the sample variance.
Sample variance formula is as follows.
\(s^2=\frac{1}{n−1}\sum_{i=1}^n(x_i−\overline{x})^2\)
Where:
\(s^2\)=Sample Variance.
n= Number of observations in the sample.
\(x_{i}\)=ith observation in the sample.
\(\overline{x}\)=Sample mean or Arithmetic mean.
The extent of the variance corresponds to the dimension of the overall range of numbers; meaning the variance is higher when there is a broader range of numbers in the group, and the variance is smaller when there is a narrower array of numbers.
Below is the formula for standard deviation regarding population and sample:
Population standard deviation is given by the formula:
\(σ=\sqrt{\frac{1}{N}\sum_{i=1}^N(X_i−μ)^2}\)
Where:
\(\sigma\)= Population standard deviation
With samples, we practice ‘n – 1’ in the formula because applying ‘n’ would provide us with a biased estimate that consistently minimizes variability. The sample variance would be lower than the actual variance of the population.
Sample standard deviation formula is as follows.
\(s=\sqrt{\frac{1}{n−1}\sum_{i=1}^n(x_i−\overline{x})^2}\)
Where:
s = Sample standard deviation
Reducing the sample ‘n’ to ‘n – 1’ gets the variance artificially large, providing you with an unbiased estimate of variability. We practice such a model as it is better to overestimate rather than underestimate variability in samples given.
Learn about Mean Deviation
Now that you know the variance is the square of standard deviation as per the formula and definition, let us step towards how to calculate variance and standard deviation and also learn how to find standard deviation from variance with a basic example.
Consider the given sample space where n = 5 and the data set is given as = { 1,2,3,4,5}.
Standard deviation and variance are 2 different mathematical theories that are both closely correlated to one another. The variance is required to calculate the standard deviation. These numbers help dealers and investors define the volatility of an expense and hence allow them to make well-informed trading decisions.
Variance is equivalent to the average squared deviations from the mean, while standard deviation implies the number’s square root. The standard deviation is a square root of the obtained variance.
Both models exhibit variability in distribution, although their units vary: i,e the standard deviation is represented in the same units as the primary values, whereas the variance is represented in squared units.
In statistics, variance and standard deviation are two important measures that show how spread out the data is from the mean (average). They help us understand whether the numbers are close to the average or spread far away.
Where:
Suppose we have 5 test scores: 2, 4, 6, 8, 10
Step 1: Find the mean (x̄)
Mean = (2 + 4 + 6 + 8 + 10) ÷ 5 = 30 ÷ 5 = 6
Step 2: Find each deviation (xi − x̄)
Step 3: Square each deviation
Step 4: Find variance
Variance = (16 + 4 + 0 + 4 + 16) ÷ 5 = 40 ÷ 5 = 8
Step 5: Find standard deviation
Standard Deviation = √8 ≈ 2.83
Some important points on variance and standard deviation are as follows:
Variance and standard deviation are both ways to measure how spread out data is. Variance shows the average of the squared differences from the mean, while standard deviation is the square root of variance, making it easier to understand because it uses the same unit as the data.
|
Feature |
Variance |
Standard Deviation |
|
Meaning |
Variance tells us how much the data values are spread out from the mean (average). |
Standard deviation shows the spread of data but in the same units as the data itself, making it easier to understand. |
|
Formula |
Variance = (Sum of squared differences from the mean) ÷ (Number of values) |
Standard Deviation = Square root of Variance |
|
Unit |
The unit of variance is the square of the original data’s unit. For example, if data is in metres, variance is in square metres (m²). |
The unit of standard deviation is the same as the data’s unit. For example, if data is in metres, standard deviation is also in metres (m). |
|
Interpretation |
It gives a mathematical idea of spread, but values are not easy to compare directly with the data. |
It is easier to interpret because it is in the same unit as the data, so we can directly relate it to the numbers. |
|
Value |
Always non-negative, and usually larger than the standard deviation. |
Always non-negative, and is the square root of variance, so usually smaller. |
|
Usefulness |
Good for statistical calculations and theory. |
Good for practical understanding and real-life comparison of data spread. |
Let’s learn how to calculate variance and standard deviation with some solved examples from the examination viewpoint as well.
Solved Example 1: The standard deviation of 4, 6, 10, 5, 10, is?
Solution:
Here, n = 5,
Mean (x̅) =\(\sum_{i=1}^5\frac{x_i}{5}=\frac{\left(4+6+10+5+10\right)}{5}=\frac{35}{5}=7\)
Variance
\((σ^2)=\sum_{i=1}^n\frac{\left(x_i−\mu\right)^2}{n}=\frac{(4−7)^2+(6−7)^2+(10−7)^2+(5−7)^2+(10−7)^2}{5}\)
\(=\frac{9+1+9+4+9}{5}=6.4\)
Standard deviation \((σ) = \sqrt{σ^2} = \sqrt{6.4}\)
Solved Example 2: If the standard deviation of a distribution is 29. Find the variance?
Solution:
\(\text{Variance}=(S.D)^2\)
\(\text{Variance}=(29)^2=841\)
Solved Example 3: For the given data set 1, 1, 2, 3, 1, 1, 12, 1, 5 find the value of variance?
Solution: Variance=\(σ^{2}=\frac{1}{N}\sum_{i=1}^N(X_i−μ)^{2}\)
To start with obtaining the mean first;
Mean = (1+ 1+ 2+ 3+ 1+ 1+ 12+ 1+ 5)/9 = 27/9 = 3
Mean Deviation (M.D) about mean= Subtract mean from each given data set values.
\(\left(X_i−μ\right)^2=(2^2+2^2+1^2+0^2+2^2+2^2+9^2+2^2+2^2)=106\)
Substituting the values in the formula:
The value of Variance = \(\frac{106}{9}=11.77\)
Solved Example 4: If the mean and the coefficient variation of distribution is 25% and 35% respectively, find variance.
Solution: The relation between mean, coefficient of variation and the standard deviation is as follows:
\(\text{Coefficient of variation}=\frac{\text{S.D}}{\text{Mean}}\times100\)
⇒\(35=\frac{\text{S.D}}{25}\times100\)
⇒ S.D = 8.75
\(\text{Variance}=(SD)^2\)
= \((8.75)^2 = 76.5625\)
We hope that the above article on Variance and Standard Deviation is helpful for your understanding and exam preparations. Stay tuned to the Testbook App for more updates on related topics from Mathematics, and various such subjects. Also, reach out to the test series available to examine your knowledge regarding several exams.
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