The variance of the first 50 even natural num | vedclass

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(B) The first $50$ even natural numbers are $2, 4, 6, \ldots, 100$. Mean,$\bar{x} = \frac{\sum_{i=1}^{50} x_i}{50} = \frac{2(1+2+3+\ldots+50)}{50} = \

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$833$

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(B) The first $50$ even natural numbers are $2, 4, 6, \ldots, 100$. Mean,$\bar{x} = \frac{\sum_{i=1}^{50} x_i}{50} = \frac{2(1+2+3+\ldots+50)}{50} = \frac{2 	imes \frac{50 	imes 51}{2}}{50} = 51$. Variance,$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. $\sum x_i^2 = 2^2 + 4^2 + \ldots + 100^2 = 4(1^2 + 2^2 + \ldots + 50^2)$. Using the sum of squares formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get: $\sum x_i^2 = 4 	imes \frac{50(51)(101)}{6} = \frac{2 	imes 50 	imes 51 	imes 101}{3} = 2 	imes 50 	imes 17 	imes 101 = 171700$. $\sigma^2 = \frac{171700}{50} - (51)^2 = 3434 - 2601 = 833$.

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

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