JEE Main 2025 — Statistics Question with Solution

$$\begin{array}{rl}& \sum _{\mathrm{i}=1}^{10}{\mathrm{x}}_{\mathrm{i}}=50,\therefore \text{ mean }=5\\ & \text{ Variance }=\frac{4}{5}=\frac{\sum {\mathrm{x}}_{\mathrm{i}}^2}{10}-{\left(\frac{\sum {\mathrm{x}}_{\mathrm{i}}}{10}\right)}^2\\ & \frac{4}{5}=\frac{\sum {\mathrm{x}}_{\mathrm{i}}^2}{10}-25\\ & \Rightarrow \sum {\mathrm{x}}_{\mathrm{i}}^2=258\text{.... (1)}\\ & \text{ Now }\sum _{\mathrm{i}=1}^{10}{\left({\mathrm{x}}_{\mathrm{i}}-\beta \right)}^2=98\\ & \sum _{\mathrm{i}=1}^{10}\left({\mathrm{x}}_{\mathrm{i}}^2-2\beta \cdot {\mathrm{x}}_{\mathrm{i}}+{\beta }^2\right)=98\\ & 258-2\beta (50)+10{\beta }^2=98\\ & (\beta -8)(\beta -2)=0\\ & \beta =\text{ or }\beta =2(\text{ as }\beta >2)\\ & \therefore \beta =8\text{..... (2)}\end{array}$$

Now as per the question

$$\begin{array}{rl}& 2\left({\mathrm{x}}_1-1\right)+4\beta ,2\left({\mathrm{x}}_2-1\right)+4\beta ,\dots .2\left({\mathrm{x}}_{10}-1\right)+4\beta \\ & \text{ can be simplified to }\\ & 2{\mathrm{x}}_1+30,2{\mathrm{x}}_2+30,\dots ..2{\mathrm{x}}_{10}+30\text{ using eq. (2) }\\ & \mu =2(5)+30=40\text{..... (3)}\\ & {\sigma }^2=2^2\left(\frac{4}{5}\right)=\frac{16}{5}\\ & \because \frac{\beta \mu }{{\sigma }^2}=\frac{8\times 40}{16/5}=100\end{array}$$