Question

The variance of the first $20$ positive integral multiples of $4$ is equal to

• A. $532$
• B. $133$
• C. $266$
• D. $600$

Answer 1

Answer: (A)

The numbers are $4,8,12........80$
Mean, $\left(\right.\bar{x}\left.\right)=\frac{4 \left(1 + 2 + 3 + . . . . . . 20\right)}{20}=42$
$\Sigma x_{i}^{2}=4^{2}\left(1^{2} + 2^{2} + \left(. . . . . . 20\right)^{2}\right)$
$=\frac{16 \times 20 \times 21 \times 41}{6}=45920$
So, the variance is $=\frac{\Sigma x_{i}^{2}}{20}-\left(\right. \overset{-}{x} \left.\right)^{2}=2296-1764=532$