If both mean and variance of 50 observations $x_1, x_2, \ldots, x_{50}$ 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

Question

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Mean }=\frac{\sum x_i}{n} \ & \Rightarrow 16=\frac{\sum x_{\mathrm{i}}}{50} \ & \Rightarrow \sum x_{\mathrm{i}}=800 \ & 	ext { Variance }=\frac{\sum x_{\mathrm{i}}^2}{\mathrm{n}}-(\bar{x})^2 \ & \Rightarrow 256=\frac{\sum x_{\mathrm{i}}^2}{50}-(16)^2 \ & \Rightarrow \frac{\sum x_{\mathrm{i}}^2}{50}=512 \ & \Rightarrow \sum x_{\mathrm{i}}^2=25600 \ & 	ext { New mean }=\frac{\sum\left(x_{\mathrm{i}}-5ight)^2}{\mathrm{n}} \ & =\frac{\sum x_{\mathrm{i}}^2-10 \sum x_{\mathrm{i}}+\sum 25}{50} \ & =\frac{25600-10(800)+25(50)}{50} \ & =\frac{18850}{50} \ & =377 \ & \end{aligned}$

If both mean and variance of 50 observations $x_1, x_2, \ldots, x_{50}$ are equal to 16 and 256 respectively…

Solution