The standard deviation of the first n odd nat | vedclass

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(B) The first $n$ odd natural numbers are $1, 3, 5, \dots, (2n - 1)$.The sum of these numbers is $\sum x_i = n^2$.The mean is $\bar{x} = \frac{\sum x_

Vedclass · Accepted Answer

$\sqrt{\frac{n^2 - 1}{3}}$

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(B) The first $n$ odd natural numbers are $1, 3, 5, \dots, (2n - 1)$.The sum of these numbers is $\sum x_i = n^2$.The mean is $\bar{x} = \frac{\sum x_i}{n} = \frac{n^2}{n} = n$.The sum of squares is $\sum x_i^2 = 1^2 + 3^2 + 5^2 + \dots + (2n - 1)^2$.Using the formula for the sum of squares of the first $n$ odd numbers: $\sum_{k=1}^{n} (2k-1)^2 = \frac{n(2n-1)(2n+1)}{3} = \frac{n(4n^2-1)}{3}$.The variance $\sigma^2$ is given by $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.$\sigma^2 = \frac{n(4n^2-1)}{3n} - n^2 = \frac{4n^2-1}{3} - n^2$.$\sigma^2 = \frac{4n^2 - 1 - 3n^2}{3} = \frac{n^2 - 1}{3}$.Therefore,the standard deviation $\sigma = \sqrt{\frac{n^2 - 1}{3}}$.

The standard deviation of the first $n$ odd natural numbers is ..........

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