Difference Between Variance and Standard Deviation

Variance and Standard deviation are both widely used in mathematics to solve statistical 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.

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.

Formula of Standard Deviation

There are two formulas for Standard Deviation:

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 the data set is dispersed around its mean.
• It is simply the square of the standard deviation and is denoted as σ².
• Its unit is different from that of the individual items in the statistical series and, hence, it is less suitable for direct comparison as a measure of dispersion.

Formula for Variance

There are two formula for Variance:

Variance vs Standard Deviation

The basic difference between Variance and Standard Deviation is discussed in the table below,

Also Check:

• Standard Deviation Formula
• Variance and Standard Deviation
• Sample Variance vs Population Variance
• Measures of spread

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 = \Sigma\dfrac{(X_i-\bar{X})^2}{N}\\=\frac{68}{8}     

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 = \sum x^2=\frac{n(n+1)(2n+1)}{6}

Now, standard deviation = σ=\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}}

Thus the standard deviation of first n natural numbers is \sqrt{\frac{n^2-1}{12}}   .

Example 3: Find the standard deviation of the following data:

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 = \Sigma\dfrac{(X_i-\bar{X})^2}{N}     = 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)=∑Z_i . P(Z_i)  = (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

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 = \Sigma\dfrac{f(M-\bar{X})^2}{N}

Population Variance = 62.98

Example 6: Find the variance of the first 69 natural numbers.

Solution:

Standard deviation of first n natural numbers is \sqrt{\frac{n^2-1}{12}}   .

Thus, Variance = {n²-1}/12

Here, n = 69

Thus, 

Var = {69²-1}/12

Var = 396.67

Practice Problems

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