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

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

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): $$\sigma = \sqrt{\sigma^2}$$​​

​