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Find the standard deviation of the first ‘n’ natural numbers.

Question

Find the standard deviation of the first ‘n’ natural numbers.

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

$\frac{n(2n+1)}{3}$

$\frac{n(n+1)}{12}$

$\sqrt{\frac{n^2+1}{6}}$

Solution

Correct option is A

Given

The first n natural numbers are 1, 2, 3,……,n.

Formula Used

Mean (μ) of the first n natural numbers:

\mu = \frac{\text{Sum of first } n \text{ natural numbers}}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}

Variance (

\sigma^2

) of the first n natural numbers:

\sigma^2 = \frac{\text{Sum of squares of first } n \text{ natural numbers}}{n} - \mu^2

\text{Sum of squares} = \frac{n(n+1)(2n+1)}{6}

Standard deviation (σ):

\sigma = \sqrt{\sigma^2}

Solution

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Find the standard deviation of the first ‘n’ natural numbers.

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

$\frac{n(2n+1)}{3}$

$\frac{n(n+1)}{12}$

$\sqrt{\frac{n^2+1}{6}}$

Correct option is A

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Find the standard deviation of the first ‘n’ natural numbers.

​n2−112\sqrt{\frac{n^2 -1}{12}}12n2−1​​​​

​n(2n+1)3\frac{n(2n+1)}{3}3n(2n+1)​​​

​n(n+1)12\frac{n(n+1)}{12}12n(n+1)​​​

​n2+16\sqrt{\frac{n^2+1}{6}}6n2+1​​​​

Correct option is A

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

$\frac{n(2n+1)}{3}$

$\frac{n(n+1)}{12}$

$\sqrt{\frac{n^2+1}{6}}$