Mathematics - Statistics Question with Solution | TestHub

Let the observations be $x_1,x_2,\dots ,x_{20}$ 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 $y_i=2x_i$ 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 $=\frac{1}{20}\times 400=20=2^2\times 5$.