Let $x_1, x_2, ..., x_{10}$ be ten observations such that $\sum\limits_{i=1}^{10} (x_i - 2) = 30$, $\sum\limits_{i=1}^{10} (x_i - \beta)^2 = 98$, $\beta > 2$, and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of $2(x_1 - 1) + 4\beta, 2(x_2 - 1) + 4\beta, ..., 2(x_{10} - 1) + 4\beta$, then $\frac{\beta\mu}{\sigma^2}$ is equal to :

<p>$$\begin{aligned} & \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}=50, \quad \therefore \text { mean }=5 \\ & \text { Variance }=\frac{4}{5}=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{10}-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{10}\right)^2 \\ & \frac{4}{5}=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{10}-25 \\ & \Rightarrow \sum \mathrm{x}_{\mathrm{i}}^2=258 \quad\text{.... (1)}\\ & \text { Now } \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-\beta\right)^2=98 \\ & \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}^2-2 \beta \cdot \mathrm{x}_{\mathrm{i}}+\beta^2\right)=98 \\ & 258-2 \beta(50)+10 \beta^2=98 \\ & (\beta-8)(\beta-2)=0 \\ & \beta=\text { or } \beta=2 \quad(\text { as } \beta>2) \\ & \therefore \beta=8\quad\text{..... (2)} \end{aligned}$$</p> <p>Now as per the question</p> <p>$$ \begin{aligned} &2\left(\mathrm{x}_1-1\right)+4 \beta, 2\left(\mathrm{x}_2-1\right)+4 \beta, \ldots .2\left(\mathrm{x}_{10}-1\right)+4 \beta\\ &\text { can be simplified to }\\ &2 \mathrm{x}_1+30,2 \mathrm{x}_2+30, \ldots . .2 \mathrm{x}_{10}+30 \text { using eq. (2) }\\ &\mu=2(5)+30=40\quad\text{..... (3)}\\ &\sigma^2=2^2\left(\frac{4}{5}\right)=\frac{16}{5}\\ &\because \frac{\beta \mu}{\sigma^2}=\frac{8 \times 40}{16 / 5}=100 \end{aligned}$$</p>