The variance of the first 50 even natural num | vedclass

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Question

(A) The first $50$ even natural numbers are $2, 4, 6, \dots, 100$. The mean $\bar{x} = \frac{2+4+\dots+100}{50} = \frac{2(1+2+\dots+50)}{50} = \frac{2

Vedclass · Accepted Answer

$833$

Vedclass · Answer

(A) The first $50$ even natural numbers are $2, 4, 6, \dots, 100$. The mean $\bar{x} = \frac{2+4+\dots+100}{50} = \frac{2(1+2+\dots+50)}{50} = \frac{2 	imes \frac{50 	imes 51}{2}}{50} = 51$. The variance is given by $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. $\sum x_i^2 = 2^2 + 4^2 + \dots + 100^2 = 4(1^2 + 2^2 + \dots + 50^2)$. Using the sum of squares formula $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$: $\sum x_i^2 = 4 	imes \frac{50 	imes 51 	imes 101}{6} = 2 	imes 50 	imes 17 	imes 101 = 171700$. $\sigma^2 = \frac{171700}{50} - (51)^2 = 3434 - 2601 = 833$.

The variance of the first $50$ even natural numbers is

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