Question:
If both the mean and the standard deviation of 50 observations $\mathrm{x}_{1}, \mathrm{x}_{2} \ldots, \mathrm{x}_{50}$ are equal to 16 , then the mean of $\left(x_{1}-4\right)^{2},\left(x_{2}-4\right)^{2}, \ldots . .\left(x_{50}-4\right)^{2}$ is :
Correct Option: , 4
Solution:
$\operatorname{Mean}(\mu)=\frac{\sum \mathrm{x}_{\mathrm{i}}}{50}=16$
standard deviation $(\sigma)=\sqrt{\frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{50}-(\mu)^{2}}=16$
$\Rightarrow(256) \times 2=\frac{\sum x_{i}^{2}}{50}$
$\Rightarrow$ New mean
$=\frac{\sum\left(x_{i}-4\right)^{2}}{50}=\frac{\sum x_{i}^{2}+16 \times 50-8 \sum x_{i}}{50}$
$=(256) \times 2+16-8 \times 16=400$