If A and B are the variances of the first n e | vedclass

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(A) The variance of the first $n$ natural numbers is given by $\sigma^2 = \frac{n^2-1}{12}$.For the first $n$ even numbers $(2, 4, 6, \dots, 2n)$,each

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$A=B$

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(A) The variance of the first $n$ natural numbers is given by $\sigma^2 = \frac{n^2-1}{12}$.For the first $n$ even numbers $(2, 4, 6, \dots, 2n)$,each term is $2$ times the corresponding natural number. Thus,the variance $A = 2^2 	imes \frac{n^2-1}{12} = \frac{4(n^2-1)}{12} = \frac{n^2-1}{3}$.For the first $n$ odd numbers $(1, 3, 5, \dots, 2n-1)$,these are obtained by subtracting $1$ from each of the first $n$ even numbers. Since variance is invariant under change of origin,the variance $B$ of the first $n$ odd numbers is the same as the variance of the first $n$ even numbers.Therefore,$A = B$.

If $A$ and $B$ are the variances of the first $n$ even numbers and the first $n$ odd numbers respectively,then:

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