Let $x_1, x_2, \ldots, x_{10}$ be ten observations such that $\sum_{i=1}^{10}\left(x_i-2\right)=30, \sum_{i=1}^{10}\left(x_i-\beta\right)^2=98, \beta \gt 2$, and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of \(2\left(x_1-1\right)+4…

Question

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

Aicharya · Accepted Answer

$\begin{aligned}
& \sum_{l=1}^{10}\left(x_l-2\right)=30 \\ & \sum_{i=1}^{10} x_l=50 \\ & \Rightarrow \text { Mean }=5 \\ & \text { Variance }=\frac{4}{5}=\frac{\sum x_l^2}{10}-(\bar{x})^2 \\ & \frac{4}{5}=\frac{\sum x_l^2}{10}-25 \\ & \Rightarrow \sum x_l^2=258
\end{aligned}$
Now, $\sum_{l=1}^{10}\left(x_l-\beta\right)^2=98$
$\begin{aligned}
& \sum_{l=1}^{10} x_l^2-2 \beta \sum_{l=1}^{10} x_l+10 \beta^2=98 \\ & \Rightarrow 258-2 \beta(50)+10 \beta^2=98 \\ & \Rightarrow 10 \beta^2-100 \beta+160=0 \\ & \Rightarrow \beta^2-10 \beta+16=0 \\ & \Rightarrow \beta=8 \text { as } \beta>2
\end{aligned}$
Now, as per the question
$2\left(x_1-1\right)+4 \beta, 2\left(x_2-1\right)+4 \beta, \ldots 2\left(x_{10}-1\right)+4 \beta$
Can be simplified as
$2 x_1+30,2 x_2+30, \ldots, 2 x_{10}+30$
$\begin{aligned} & \mu=2(5)+30=40 \\ & \sigma^2=2^2\left(\frac{4}{5}\right)=\frac{16}{5} \\ & \frac{\beta \mu}{\sigma^2}=\frac{8 \times 40}{\frac{16}{5}}=100\end{aligned}$ ^

Let \(x_1, x_2, \ldots, x_{10}\) be ten observations such that \(\sum_{i=1}^{10}\left(x_i-2\right)=30,…

Solution