Difference Between Variance and Standard Deviation
Last Updated :
15 Apr, 2025
Variance and Standard deviation both formulas are widely used in mathematics to solve statistics problems. They provide various ways to extract information from the group of data.
They are also used in probability theory and other branches of mathematics. So it is important to distinguish between them. Let’s learn about Standard Deviation, Variance, and their difference in detail in this article.
What is Standard Deviation?
Standard deviation is an important method of measuring statistical deviation. It is the measure of the extent to which the numbers in a statistical series are spread from their arithmetic mean. It is always a non-negative value, and its unit is the same as that of the items in the given series. The same unit among both makes comparison and interpretation easier and more detailed. It is a common tool used to measure central tendency and is denoted as σ and is given as:
Formula of Standard Deviation
There are two formulas for Standard Deviation:
- Population Standard Deviation

Population Standard Deviation Formula, Image Credit – arwaaalkindi (gfg author)
In most cases, the Population Standard Deviation Formula is used:
S.D.(x) = σ =[Tex]\sqrt{\frac{1}{N} \sum_{i=1}^{N}\left(X_{i}-\mu\right)^{2}}[/Tex]
- Sample Standard Deviation

Sample Standard Deviation Formula, Image Credit – arwaaalkindi (gfg author)
What is Variance?
Variance is the squared deviation of items/values in a statistical series from its arithmetic mean. This numerical value quantifies the average magnitude to which extent the data set is dispersed around itself. It is simply the square of the standard deviation and is denoted as σ2. Since it is the squared deviation, its unit is different from that of the individual items in the statistical series and is, hence, not a reliable measure of dispersion because of inaccurate and vague comparison. Variance is calculated as
Formula for Variance
There are two formula for Variance:

Population Variance Formula, Credit-arwaaalkindi (gfg author)
In most cases, the Population Variance Formula isis used:
Var(x) = [Tex]\sum \frac{f\left ( M_{i}-\overline{X} \right )^{2}}{N}[/Tex]

Sample Variance Formula, Credit-arwaaalkindi (gfg author)
Difference Between Variance and Standard Deviation
The basic difference between Variance and Standard Deviation is discussed in the table below,
Standard Deviation | Variance |
---|
It is the measure of the dispersion of values in a given data set relative to their mean. | It is the statistical measure of how far the numbers are spread in a data set from their average. |
It measures the absolute variability of the dispersion. | It helps determine the size of the data spread. |
It is calculated by taking the square root of the variance. | It is calculated by taking the average of the squared deviations of each value from the average. |
It is primarily used as a measure of market and security volatility in finance. | It is one of the key aspects of asset allocation in investing portfolios. |
Also Read:
Solved Examples on Variance and Standard Deviation
Example 1: Find the variance and standard deviation of the following data
Solution:
X
| x = X – X̄
| x2
|
---|
5
| -1
| 1
|
8
| 2
| 4
|
3
| -3
| 9
|
6
| 0
| 0
|
7
| 1
| 1
|
12
| 6
| 36
|
5
| -1
| 1
|
2
| -4
| 16
|
Σx = 48
|
| Σx2 = 68
|
Arithmetic Mean = Σx/N = 48/8 = 6
Population Variance = [Tex]\Sigma\dfrac{(X_i-\bar{X})^2}{N}\\=\frac{68}{8} [/Tex]
Population Variance = 8.5
Standard Deviation = √8.5 = 2.91
Example 2: Find the standard deviation of first n natural numbers.
Solution:
Mean of first n natural numbers = Sum of first n natural numbers/n
= [n(n+1)/2/]n
= (n+1)/2
Sum of squares of first n natural numbers = [Tex]\sum x^2=\frac{n(n+1)(2n+1)}{6}[/Tex]
Now, standard deviation = σ=[Tex]\sqrt{\frac{\sum x^2} N-(\frac{\sum x} N)^2}\\=\sqrt{\frac{n(n+1)(2n+1)}{6}-(\frac{n+1}2)^2}\\=\sqrt{\frac{n^2-1}{12}}[/Tex]
Thus the standard deviation of first n natural numbers is [Tex]\sqrt{\frac{n^2-1}{12}} [/Tex].
Example 3:Find the standard deviation of the following data:
X
| 5
| 10
| 15
| 20
| 25
| 30
|
f
| 6
| 7
| 3
| 2
| 1
| 1
|
Solution:
X
| f
| fX
| x = X – X̄
| x2
| fx2
|
5
| 6
| 30
| -7
| 49
| 294
|
10
| 7
| 70
| -2
| 4
| 28
|
15
| 3
| 45
| 3
| 9
| 27
|
20
| 2
| 40
| 8
| 64
| 128
|
25
| 1
| 25
| 13
| 169
| 169
|
30
| 1
| 30
| 18
| 324
| 324
|
| N = 20
| ΣfX = 240
|
|
| Σfx2 = 970
|
Arithmetic Mean = ΣfX/N = 240/20 = 12
Population Variance = [Tex]\Sigma\dfrac{(X_i-\bar{X})^2}{N} [/Tex] = 970/20
Population Variance = 48.5
Thus, Standard Deviation = √48.5 = 6.96
Example 4: Find the variance of the variable Z, where Z represents the sum of all observations upon rolling a pair of dice.
Solution:
There are total 36 observations when a pair of dice is rolled. The probabilities of getting different sums upon rolling a pair of dice are:
P(Z=2) = 1/36
P(Z=3) = 2/36 = 1/18
P(Z=4) = 3/36 = 1/12
P(Z=5) = 4/36 = 1/9
P(Z=6) = 5/36
P(Z=7) = 6/36 = 1/6
P(Z=8) = 5/36
P(Z=9) = 1/9
P(Z=10) = 1/12
P(Z=11) = 1/18
P(Z=12) = 1/36
Since, E(Z)=∑Zi . P(Zi) = (1/18) + (1/6) + (1/3) + (5/9) + (5/6) + (7/6) + (10/9) + 1 + (5/6) + (11/18) + (1/3)
Thus, E(Z) = 7
And, E(Z2) = 54.833
Thus,
Var(Z) = E(Z2) – E(Z)2
= 54.833 – 49
Var(Z) = 5.833
Example 5: Calculate variance for the following grouped data
Intervals | 20 – 25 | 25 – 30 | 30 – 35 | 35 – 40 | 40 – 45 | 45 – 50 |
Frequency | 170 | 110 | 80 | 45 | 40 | 35 |
Solution:
X
| f
| M
| d
| d2
| fd
| fd2
|
20 – 25
| 170
| 22.5
| -2
| 4
| -340
| 680
|
25 – 30
| 110
| 27.5
| -1
| 1
| -110
| 110
|
30 – 35
| 80
| 32.5
| 0
| 0
| 0
| 0
|
35 – 40
| 45
| 37.5
| 1
| 1
| 45
| 45
|
40 – 45
| 40
| 42.5
| 2
| 4
| 80
| 160
|
45 – 50
| 35
| 47.5
| 3
| 9
| 105
| 315
|
| N = 480
|
|
|
| Σfd = -220
| Σfd2 = 1340
|
Variance = [Tex]\Sigma\dfrac{f(M-\bar{X})^2}{N}[/Tex]
Population Variance = 62.98
Example 6: Find the variance of the first 69 natural numbers.
Solution:
Standard deviation of first n natural numbers is [Tex]\sqrt{\frac{n^2-1}{12}} [/Tex].
Thus, Variance = {n2-1}/12
Here, n = 69
Thus,
Var = {692-1}/12
Var = 396.67
Practice Problems (Try and master)
Question 1 : Find the variance and standard deviation of all the possibilities of rolling a die.
Question 2 : Find the variance and standard deviation of all the even numbers less than 10.
Question 3 : Find the standard deviation for the data: 42, 38, 35, 26, 45, 52, 48.
Answer 1 : Mean, x̅ = 3.5
Variance (σ²) = 2.917
Standard Deviation (σ) = √(2.917) = 1.708
Answer 2 : Mean, x̅ = 4
Variance (σ²) = 8
Standard Deviation (σ) = √(8) = 2.828
Answer 3 : Mean, x̅ = 40.85
Standard Deviation (σ) = 8.07
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