Question

If both mean and the standard deviation of $50$ observation $x_1,x_2,...,x_{50}$ are equal to $16$, then the mean of $(x_1-4{)}^2,(x_2-4{)}^2,...,(x_{50}-4{)}^2$ is :

• A. $380$
• B. $400$
• C. $480$
• D. $525$

Answer 1

The correct option is B $400$

Mean$\left(\mu \right)=\frac{\sum x_i}{50}=16$
$\Rightarrow \sum x_i=16\times 50$

Standard deviation$=\left(\sigma \right)=\sqrt{\frac{\sum x_i^2}{50}-(\mu {)}^2}=16$
$\Rightarrow \frac{\sum x_i^2}{50}={16}^2+{16}^2=512$
$\Rightarrow \sum x_i^2=512\times 50$

Required mean $=\frac{\sum (x_i-4{)}^2}{50}$
$=\frac{\sum x_i^2+16\times 50-8\sum x_i}{50}$
$=512+16-8\times 16$
$=400$