The variance of the first 50 even natural num | vedclass

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(D) The first $50$ even natural numbers are $2, 4, 6, \dots, 100$.The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.Here,$n

Vedclass · Accepted Answer

$833$

Vedclass · Answer

(D) The first $50$ even natural numbers are $2, 4, 6, \dots, 100$.The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.Here,$n = 50$. The sum of the first $n$ even numbers is $n(n+1) = 50 	imes 51 = 2550$. Thus,the mean $\bar{x} = \frac{2550}{50} = 51$.The sum of squares is $\sum x_i^2 = 2^2 + 4^2 + \dots + 100^2 = 2^2(1^2 + 2^2 + \dots + 50^2)$.Using the formula $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$,we get $\sum x_i^2 = 4 	imes \frac{50(51)(101)}{6} = 4 	imes \frac{257550}{6} = 4 	imes 42925 = 171700$.Now,$\frac{\sum x_i^2}{n} = \frac{171700}{50} = 3434$.Variance $\sigma^2 = 3434 - (51)^2 = 3434 - 2601 = 833$.

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

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