Question

If the standard deviation of n elements of the series x₁, x₂, x₃ ....... , x_n is σ, then find the variance of the series ax₁, ax₂, ax₃ ....... , ax_n is:
1. a²σ  
2. a²nσ  
3. aσ 
4. a²σ²

Answer 1

Correct Answer - Option 4 : a²σ²

Concept:

Variance of a series = (Standard deviation)² 

Standard deviation = \(\rm \sqrt{\sum_{1}^{n}(x_i - \overline x)^2\over n}\)

Where x_i is the element of the series and \(\rm \overline x\) is the mean of the series having total n elements

Calculation:

Given Standard deviation of series x1, x2, x3 ....... , xn is σ

\(\rm {1\over n}\sqrt{\sum_{1}^{n}(x_i - \overline x)^2}\) = σ 

Where \(\rm \overline x\) = \(\rm x_1 + x_2 + x_3 ....... + x_n\over n\)

Variance of this series = \(\rm {1\over n}\sum_{1}^{n}(x_i - \overline x)^2\) = σ²

Now for the series ax1, ax2, ax3 ....... , axn 

Mean = \(\rm a\left(x_1 + x_2 + x_3 ....... + x_n\over n\right)\) = a\(\rm \overline x\)

Variance = \(\rm {1\over n}\sum_{1}^{n}(ax_i - a\overline x)^2\)

V = \(\rm {a^2\over n}\sum_{1}^{n}(x_i - \overline x)^2\)

V = a² ×\(\rm \left[{1\over n}\sum_{1}^{n}(x_i - \overline x)^2\right]\)

V = a²σ²