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(D) The first $n$ even natural numbers are $2, 4, 6, \dots, 2n$.The mean $\bar{x} = \frac{2(1+2+\dots+n)}{n} = \frac{2 	imes n(n+1)}{2n} = n+1$.The v

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Statement-$I$ is false and Statement-$II$ is true.

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(D) The first $n$ even natural numbers are $2, 4, 6, \dots, 2n$.The mean $\bar{x} = \frac{2(1+2+\dots+n)}{n} = \frac{2 	imes n(n+1)}{2n} = n+1$.The variance is given by $\sigma^2 = \frac{1}{n} \sum_{i=1}^n (2i)^2 - (n+1)^2$.$\sigma^2 = \frac{4}{n} 	imes \frac{n(n+1)(2n+1)}{6} - (n+1)^2 = \frac{2(n+1)(2n+1)}{3} - (n+1)^2$.$\sigma^2 = (n+1) \left[ \frac{4n+2-3n-3}{3} ight] = \frac{(n+1)(n-1)}{3} = \frac{n^2-1}{3}$.Thus,Statement-$I$ is false.For $n=20$,the mean is $20+1 = 21$.The variance is $\frac{20^2-1}{3} = \frac{399}{3} = 133$.The difference is $133 - 21 = 112$.Thus,Statement-$II$ is true.

Based on the following statements,choose the correct option.
Statement-$I:$ The variance of the first $n$ even natural numbers is $\frac{n^2-1}{4}$.
Statement-$II:$ The difference between the variance of the first $20$ even natural numbers and their arithmetic mean is $112$.

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Based on the following statements,choose the correct option.Statement-$I:$ The variance of the first $n$ even natural numbers is $\frac{n^2-1}{4}$.Statement-$II:$ The difference between the variance of the first $20$ even natural numbers and their arithmetic mean is $112$.