The variance of 20 observations is 5. If each observation is...

Let the observations be $x_1, x_2, \ldots, x_{20}$ and $\bar{x}$ be their mean. Given that, variance $=5$ and $n= 20. We know that, Variance$\left(\sigma^2\right)=\frac{1}{n} \sum_{i=1}^{20}$\left(x_i-\bar{x}\right)^2$ i.e. $5=\frac{1}{20} \sum_{i=1}^{20}$\left(x_i-\bar{x}\right)^2$or$\sum_{i=1}^{20}$\left(x_i-\bar{x}\right)^2=100$ .....(i) If each observation is multiplied by 2 and the new resulting observations are $y_i$, then $y_i=2 x_i$ i.e., $x_i=\frac{1}{2} y_i$ Therefore, $\overline{\mathrm{y}}=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{20} \mathrm{y}_{\mathrm{i}}=\frac{1}{20} \sum_{\mathrm{i}=1}^{20} 2 \mathrm{x}_{\mathrm{i}}=2 \cdot \frac{1}{20} \sum_{\mathrm{i}=1}^{20} \mathrm{x}_{\mathrm{i}}$ i.e., $\bar{y}=2 \bar{x}$ or $\bar{x}=\frac{1}{2} \bar{y}$. On substituting the values of $x_i$ and $\bar{x}$ in eq. (i), we get $\sum_{i=1}^{20}$\left(\frac{1}{2} y_i-\frac{1}{2} \bar{y}\right)^2=100$i.e.,$\sum_{i=1}^{20}$\left(y_i-\bar{y}\right)^2=400$ Thus, the variance of new observations $=\frac{1}{20} \times 400=20=2^2 \times 5$.