JEE Main 2025 — Statistics Question with Solution

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