Mathematics - Statistics Question with Solution | TestHub

$\begin{array}{rl}& \sum _{l=1}^{10}\left(x_l-2\right)=30\\ & \sum _{i=1}^{10}{x}_l=50\\ & \Rightarrow \text{ Mean }=5\\ & \text{ Variance }=\frac{4}{5}=\frac{\sum x_l^2}{10}-(\bar{x}{)}^2\\ & \frac{4}{5}=\frac{\sum x_l^2}{10}-25\\ & \Rightarrow \sum x_l^2=258\end{array}$ Now, ${\sum }_{l=1}^{10}{\left(x_l-\beta \right)}^2=98$ $\begin{array}{rl}& \sum _{l=1}^{10}{x}_l^2-2\beta \sum _{l=1}^{10}{x}_l+10{\beta }^2=98\\ & \Rightarrow 258-2\beta (50)+10{\beta }^2=98\\ & \Rightarrow 10{\beta }^2-100\beta +160=0\\ & \Rightarrow {\beta }^2-10\beta +16=0\\ & \Rightarrow \beta =8\text{ as }\beta >2\end{array}$ Now, as per the question $2\left(x_1-1\right)+4\beta ,2\left(x_2-1\right)+4\beta ,\dots 2\left(x_{10}-1\right)+4\beta$ Can be simplified as $2x_1+30,2x_2+30,\dots ,2x_{10}+30$ $\begin{array}{rl}& \mu =2(5)+30=40\\ & {\sigma }^2=2^2\left(\frac{4}{5}\right)=\frac{16}{5}\\ & \frac{\beta \mu }{{\sigma }^2}=\frac{8\times 40}{\frac{16}{5}}=100\end{array}$