Statement-1: The variance of the first n even | vedclass

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(B) For Statement-$1$: 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 \cdot n(n+1)

Vedclass · Accepted Answer

Statement-$1$ is true,Statement-$2$ is true. Statement-$2$ is not a correct explanation for Statement-$1$.

Vedclass · Answer

(B) For Statement-$1$: The first $n$ even natural numbers are $2, 4, 6, \dots, 2n$.The mean $\bar{x} = \frac{2(1+2+\dots+n)}{n} = \frac{2 \cdot n(n+1)}{2n} = n+1$.The variance $\sigma^2 = \frac{1}{n} \sum_{i=1}^n (2i)^2 - (n+1)^2 = \frac{4}{n} \cdot \frac{n(n+1)(2n+1)}{6} - (n+1)^2$.$\sigma^2 = \frac{2(n+1)(2n+1)}{3} - (n+1)^2 = (n+1) [\frac{4n+2-3n-3}{3}] = \frac{(n+1)(n-1)}{3} = \frac{n^2-1}{3}$.Thus,Statement-$1$ is true.For Statement-$2$: The sum of the first $n$ odd natural numbers is $\sum_{k=1}^n (2k-1) = 2 \cdot \frac{n(n+1)}{2} - n = n^2+n-n = n^2$.The sum of the squares of the first $n$ odd natural numbers is $\sum_{k=1}^n (2k-1)^2 = \sum_{k=1}^n (4k^2 - 4k + 1) = 4 \cdot \frac{n(n+1)(2n+1)}{6} - 4 \cdot \frac{n(n+1)}{2} + n$.$= \frac{2n(n+1)(2n+1) - 6n(n+1) + 3n}{3} = \frac{n[2(2n^2+3n+1) - 6n - 6 + 3]}{3} = \frac{n(4n^2+6n+2-6n-3)}{3} = \frac{n(4n^2-1)}{3}$.Statement-$2$ is true,but it does not explain the variance calculation in Statement-$1$ directly. Therefore,Statement-$2$ is not the correct explanation for Statement-$1$.

Statement-$1$: The variance of the first $n$ even natural numbers is $\frac{n^2 - 1}{3}$.
Statement-$2$: The sum of the first $n$ odd natural numbers is $n^2$ and the sum of the squares of the first $n$ odd natural numbers is $\frac{n(4n^2 - 1)}{3}$.

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Statement-$1$: The variance of the first $n$ even natural numbers is $\frac{n^2 - 1}{3}$.Statement-$2$: The sum of the first $n$ odd natural numbers is $n^2$ and the sum of the squares of the first $n$ odd natural numbers is $\frac{n(4n^2 - 1)}{3}$.