Answer:
Option B
Explanation:
Here, $\overline{X}=\frac{\sum X_{i}}{n}$
=$\frac{2+4+6+8+...+100}{50}=\frac{50 \times 51}{50}=51$
[ $\because \sum{}2n=n(n+1), $ , here n=50]
Variance , $\sigma^{2} =\frac{1}{n}\sum{}Xi^{2}-(\overline{x})^{-2}$
$\sigma^{2} =\frac{1}{50}(2^{2}+4^{2}+....+100^{2})-(51)^{2}=833$
