Statement 1: The variance of the first n odd | vedclass

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Question

(D) The first $n$ odd natural numbers are $1, 3, 5, \dots, (2n-1)$.The mean $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} (2i-1) = \frac{1}{n} \cdot n^2 = n$.

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Statement $1$ is true,Statement $2$ is true; Statement $2$ is a correct explanation for Statement $1$.

Vedclass · Answer

(D) The first $n$ odd natural numbers are $1, 3, 5, \dots, (2n-1)$.The mean $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} (2i-1) = \frac{1}{n} \cdot n^2 = n$.The sum of squares is $\sum_{i=1}^{n} (2i-1)^2 = \sum (4i^2 - 4i + 1) = 4 \frac{n(n+1)(2n+1)}{6} - 4 \frac{n(n+1)}{2} + n = \frac{n(4n^2-1)}{3}$.Variance $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2 = \frac{4n^2-1}{3} - n^2 = \frac{4n^2-1-3n^2}{3} = \frac{n^2-1}{3}$.Thus,Statement $1$ is true and Statement $2$ is true. Statement $2$ provides the correct formulas used to derive Statement $1$.

Statement $1$: The variance of the first $n$ odd 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 $n$ odd 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 $\frac{n(4n^2 - 1)}{3}$.