Find the standard deviation of the first n na | vedclass

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(A) The first $n$ natural numbers are $1, 2, 3, \ldots, n$. The mean $\bar{x} = \frac{\sum x_i}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}$. The sum of sq

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$\sqrt{\frac{n^{2}-1}{12}}$

Vedclass · Answer

(A) The first $n$ natural numbers are $1, 2, 3, \ldots, n$. The mean $\bar{x} = \frac{\sum x_i}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}$. The sum of squares is $\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}$. The variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$. $\sigma^2 = \frac{(n+1)}{2} \left[ \frac{2n+1}{3} - \frac{n+1}{2} \right] = \frac{(n+1)}{2} \left[ \frac{4n+2-3n-3}{6} \right] = \frac{(n+1)(n-1)}{12} = \frac{n^2-1}{12}$. Therefore,the standard deviation $\sigma = \sqrt{\frac{n^2-1}{12}}$.

Find the standard deviation of the first $n$ natural numbers.

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