Let x_1,x_2,...,x_10 be ten observations such that _i=1^10(x_i-2)=30, _i=1^10(x_i- )^2=98,…

Question

Let ${x}_{1},{x}_{2},...,{x}_{10}$ be ten observations such that $\sum _{i=1}^{10}({x}_{i}-2)=30$, $\sum _{i=1}^{10}({x}_{i}-\beta {)}^{2}=98$, $\beta >2$, 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 :

Accepted Answer

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

Now as per the question

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

Solutions · 1