Find the standard deviation of the first n

Solution:

Given set of first $\mathrm{n}$ natural numbers

Now we have to find the standard deviation

Given first n natural numbers, we can write in table as shown below

So, the sums of these are

$\sum x_{i}=1+2+3+\cdots+n=\frac{n(n+1)}{2}$

And

$\sum x_{i}^{2}=1^{2}+2^{2}+3^{2}+\cdots+n^{2}=\frac{n(n+1)(2 n+1)}{6}$

Therefore, the standard deviation can be written as,

$\sigma=\sqrt{\frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}}$

Substituting the values we get

$\sigma=\sqrt{\frac{\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}}{\mathrm{n}}-\left(\frac{\frac{\mathrm{n}(\mathrm{n}+1)}{2}}{\mathrm{n}}\right)^{2}}$

On simplifying

$\sigma=\sqrt{\frac{n(n+1)(2 n+1)}{6 n}-\frac{n^{2}(n+1)^{2}}{4 n^{2}}}$

$\sigma=\sqrt{\frac{n(2 n+1)+1(2 n+1)}{6}-\frac{\left(n^{2}+2 n+1\right)}{4}}$

Multiplying the numerator we get

$\sigma=\sqrt{\frac{2 n^{2}+n+2 n+1}{6}-\frac{n^{2}+2 n+1}{4}}$

Taking LCM and simplifying we get

$\sigma=\sqrt{\frac{2\left(2 n^{2}+3 n+1\right)-3\left(n^{2}+2 n+1\right)}{12}}$

$\sigma=\sqrt{\frac{4 n^{2}+6 n+2-3 n^{2}-6 n-3}{12}}$

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

Hence the standard deviation of the first $n$ natural numbers is $\sqrt{\frac{n^{2}-1}{12}}$