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Q. The variance of first $50$ even natural numbers is

JEE MainJEE Main 2014Statistics

Solution:

Variance $=\frac{\Sigma x_{i}^{2}}{N}-(\bar{x})^{2} $
$\sigma^{2}= \frac{2^{2}+4^{2}+\ldots+100^{2}}{50}-\left(\frac{2+4+\ldots+100}{50}\right)^{2} $
$= \frac{4\left(1^{2}+2^{2}+3^{2}+\ldots .+50^{2}\right)}{50}-(51)^{2} $
$=4\left(\frac{50 \times 51 \times 101}{50 \times 6}\right)-(51)^{2}$
$=3434-2601 $
$\Rightarrow \sigma^{2} =833$