Variance and Standard Deviation

Statistics give us two important measures of the spread or dispersion of a set of data points: variance and standard deviation. They tell us how far the data points are deviating from the mean. To calculate variance and standard deviation, it is quite important to follow a step-by-step approach involving the formula for the mean, variance, and standard deviation. These concepts can be widely used in every field, such as economics, physics and social sciences, where data is to be analysed.

1.0What is Variance?

Variance measures how far data points are from the mean. To calculate variance and standard deviation, first compute the mean and then find the deviations from it.

Steps to Calculate Variance:

1. Find the mean of the data set. Add all the numbers in the data set and divide by the number of values.

The formula for Mean:

$\mathrm{Mean}(\mu )=\frac{\text{ Sum of all data points }}{\text{ Number of data points }}$

2. Subtract the mean from each data point to get the deviation of each point from the mean.
3. Square each of the deviations so that there are no negative numbers, and so larger deviations receive more weight.
4. Find the average of the squared deviations. This is the variance. The formula for Variance 2:

${\sigma }^2=\frac{\Sigma {\left(x_i-\mu \right)}^2}{N}$

Where  xi are the data points, is the mean, and N is the number of data points.

For example, let's say we have the data set: 4, 7, 8, 5, and 10.

1. Mean = (4 + 7 + 8 + 5 + 10) / 5 = 34 / 5 = 6.8
2. Deviations from the mean:

• (4 - 6.8) = -2.8
• (7 - 6.8) = 0.2
• (8 - 6.8) = 1.2
• (5 - 6.8) = -1.8
• (10 - 6.8) = 3.2

3. Square each deviation:

• (-2.8)² = 7.84
• (0.2)² = 0.04
• (1.2)² = 1.44
• (-1.8)² = 3.24
• (3.2)² = 10.24

4. Sum of squared deviations = 7.84 + 0.04 + 1.44 + 3.24 + 10.24 = 22.8
5. Variance = 22.8 / 5 = 4.56. Thus, the variance of this data set is 4.56

2.0What is Standard Deviation?

Standard deviation is the square root of variance, providing an intuitive measure in the same units.

Formula for Standard Deviation ():

=xi-2N = Variance

For example, if the variance of the data set is 4.56, then the standard deviation is:

=4.56 2.14

Thus, the standard deviation is approximately 2.13. This means that, on average, the data points deviate from the mean by approximately 2.13 units.

3.0Calculating the Range, Variance, and Standard Deviation

The range of a dataset is the difference between the largest and smallest values. While range, variance, and standard deviation all measure data spread, variance and standard deviation, using the mean, provide a more complete understanding of data distribution.

4.0Solved Problems

Problem 1: Find the range of the following statistical data: 2, 4, 7, 13, 45, 32, 54

Solution: 

Range = Maximum value of frequency - minimum value of frequency 

= 54 - 2 = 52

Problem 2: Calculate the Range, variation, and standard deviation of the following frequency distribution: 

Solution: 

Range = Maximum value of frequency - minimum value of frequency

= 29 - 3 = 26

$\text{ Mean }=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{1470}{70}=21$

$\text{ Variance }=\frac{\Sigma {\left(x_i-\mu \right)}^2}{N}=\frac{1042}{70}=14.86$

Standard deviation = 3.85

Problem 3: The mean of 5 observations is 4.4, and their variance is 8.24. If three of the observations are 1, 2, and 6, find the other two observations.

Solution: Let the two observations be x and y respectively. 

Mean of the observations = 4.4

Sum of all observation total terms=4.4

1+2+6+x+y5=4.4

9+x+y = 22

x+y = 13 …….(1)

Now, Variance of the observations = 8.24

xi-2N = 8.24

(1-4.4)2+(2-4.4)2+(6-4.4)2+(x-4.4)2+(y-4.4)25=8.24

(-3.4)2+(-2.4)2+(1.6)2+(x-4.4)2+(y-4.4)25=8.24

11.56+5.76+2.56+(x-4.4)2+(y-4.4)2=41.2

(x-4.4)2+(y-4.4)2=21.32  ………(2)

By equation 1 we have x+y=13 so put y = 13–x in equation 2 

(x-4.4)2+(13-x-4.4)2=21.32

(x-4.4)2+(8.6-x)2=21.32

(x² – 8.8x + 19.36) + (x² – 17.2x + 73.96) = 21.32

2x² – 26x + 93.32 = 21.32

2x²– 26x +72=0

x² – 13x + 36 = 0

x² – 9x–4x + 36 = 0 

x(x–9) – 4(x–9) = 0 

(x–4)(x–9) = 0 

x = 4 or 9 

If x =4 then y = 13–4=9

If x = 9 then y = 13–9 = 4. 

The other two observations are 4 and 9. 

Problem 4: If V is the variance and M is the mean of the first 15 natural numbers, then what is V + M² equal to?

Solution: Let the 15 natural numbers be 1 to 15. 

The Mean of first 15 natural numbers = M

We know that the sum of the first n natural numbers is given by the formula: 

i=115n(n+1)2=15162=120

Mean (M) = 120/15 = 8

The formula for variance V is: 

V = 1ni=1n(xi-M)2

V = 1n{i=1n(xi2+M2-2Mxi)}

V = 1n(i=1nxi2+nM2-2Mi=1nxi)

The Formula for the sum of squares of the first n natural numbers is: 

i=1nxi2=n(n+1)(2n+1)6

n = 15

i=1nxi2=15(16)(31)6= 1240

Now, calculate variance by putting the values in the above formula 

V = 115(1240+1582-28120)

V = 115(1240-1920+960)=115280

V = 18.67

Now V + M² = 18.67 + 8² = 18.67+64 = 82.67