$\frac{4}{5}=\frac{\Sigma x_{i}^2}{10}-\left(\frac{\Sigma x_{i}}{10}\right)^2$
$\frac{4}{5}=\frac{\Sigma x_{i}^2}{10}-25$
$\Rightarrow \Sigma x_{i}^2=258$
$\text { Now } \sum_{i=1}^{10}\left(x_{i}-\beta\right)^2=98$
$\sum_{i=1}^{10}\left(x_{i}^2-2 \beta \cdot x_{i}+\beta^2\right)=98$
$258-2 \beta(50)+10 \beta^2=98$
$(\beta-8)(\beta-2)=0$
$\beta=\text { or } \beta=2 \quad(\text { as } \beta>2)$
$\therefore \beta=8$
Now,
$=2\left(x_1-1\right)+4 \beta, 2\left(x_2-1\right)+4 \beta, \ldots .2\left(x_{10}-1\right)+4 \beta$
$=2 x_1+30,2 x_2+30, \ldots . .2 x_{10}+30$
$\mu=2(5)+30=40$
$\sigma^2=2^2\left(\frac{4}{5}\right)=\frac{16}{5}$
$\because \frac{B \mu}{2}=\frac{8 \times 40}{}=100$