QJEEM25Ma8612: JEE Main 2025 Mathematics - Class 11 JEEM25Ma8612 Solution: Let be ten observations such that , , , and their variance is . …

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

Now as per the question

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