If both mean and variance of 50 observations $x_1, x_2, \ldots \ldots, x_{50}$ are equal to 16 and 256…
If both mean and variance of 50 observations $x_1, x_2, \ldots \ldots, x_{50}$ are equal to 16 and 256 respectively, then mean of $\left(x_1-5\right)^2,\left(x_2-5\right)^2, \ldots \ldots\left(x_{50}-5\right)^2$ is
$357$
$387$
$377$
$397$
Solution
Given that $\mathrm{n}=50, \bar{x}=16$ and $\sigma_x{ }^2=256$
$\begin{array}{ll}
\therefore & \sigma_x^2=\frac{1}{\mathrm{n}}\left(\sum_{i=1}^{50} x_i^2\right)-(\bar{x})^2 \\
\therefore & 256=\frac{1}{50}\left(\sum_{i=1}^{50} x_i^2\right)-256 \\
\therefore & \frac{1}{50}\left(\sum_{i=1}^{50} x_i^2\right)=512 \\
\therefore & \sum_{i=1}^{50} x_i^2=25600
\end{array}$
$\text { Now } \begin{aligned}
\sum_{i=1}^5\left(x_i-5\right)^2 & =\sum_{i=1}^{50} x_i^2+25 \times 50-10 \sum_{i=1}^5 x_i \\
& =25600+1250-8000
\end{aligned}$
... [From (i) and (ii)]
$\leq 18850$
$\therefore \quad \text { Required Mean }=\frac{\sum_{i=1}^{50}\left(x_i-5\right)^2}{50}=\frac{18850}{50}=377$