Question

The variance of 20 observations is 5. If each observation is multiplied by 2, find the variance of the resulting observations.

 

Answer 1

Let \[x_1 , x_2 , x_3 , . . . , x_{20}\]  be  the 20 given observations.

\[\text{ Variance}  (X) = 5\]

\[\text{ Variance } (X) = {\frac{1}{20}} \times \sum \left( {x_i - X} \right)^2 = 5 (\text{ Here , is the mean of the given observations }  . )\]

Let u_1,u_2,,u₃, ..., u₂₀ be the new observations, such that

\[u_i = 2 x_i (\text{ for }  i = 1, 2, 3, . . . , 20) . . . (1)\]

\[\text{ Mean } = \bar{U} = \frac{\sum^{20}_{i = 1} u_i}{n} \]

\[ = \frac{\sum^{20}_{i = 1} 2 x_i}{20} \left[ \text{ substituting} u_i \text{ from eq (1) and taking n as }  20 \right]\]

\[ = 2 \times \frac{\sum^{20}_{i = 1}{ x_i} }{20} \]

\[ = 2 \bar{X}\]

\[u_i - \bar{U} = 2 x_i - 2 \bar{X} (\text{ for } i = 1, 2, . . . , 20)\]

\[ = 2\left( x_i - \bar{X} \right) \]

\[ \left( u_i - \bar{U} \right)^2 = \left( 2\left( x_i - \bar{X} \right) \right)^2 \left(\text{  squaring both the sides } \right)\]

\[ = 4 \left( x_i - \bar{X} \right)^2 \]

\[ \therefore \sum^{20}_{i = 1} \left( u_i - \bar{U} \right)^2 = \sum 4^{20}_{i = 1} \left( x_i - \bar{X} \right)^2 \]

\[\frac{\sum^{20}_{i = 1} \left( u_i - \bar{U} \right)^2}{20} = \frac{\sum 4^{20}_{i = 1} \left( x_i - \bar{X} \right)^2}{20}\]

\[ = 4 \frac{\sum^{20}_{i = 1} \left( x_i - \bar{X} \right)^2}{20}\]

\[\text{ Variance } (U) = 4 \times \text{ Variance }(X)\]

\[ = 4 \times 5 \]

\[ = 20\]

 Thus, variance of the new observations is 20.