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Find the standard deviation of the first n natural numbers.
CBSE | Class 11 | Maths | NCERT Exemplar | Statistics
Find the standard deviation of the first n natural numbers.

Solution:

It is given: set of first n natural numbers

We now need to find the standard deviation

Provided first n natural numbers, it can be written in table as shown below

    \[\begin{tabular}{|l|l|l|l|l|l|l|l|l|} \hline $\mathrm{x}_{\mathrm{i}}$ & 1 & 2 & 3 & 4 & 5 & $\ldots$ & $\ldots .$ & $\mathrm{n}$ \\ \hline $\mathrm{x}_{\mathrm{i}}^{2}$ & 1 & 4 & 9 & 16 & 25 & $\ldots .$ & $\ldots .$ & $\mathrm{n}^{2}$ \\ \hline \end{tabular}\]

Therefore, 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}

So, we can write the standard deviation as,

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

Substitute the values

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

Simplify

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

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

Multiplyi the numerator

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

Take LCM and simplify

\begin{aligned} &\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}} \end{aligned}

As a result, \sqrt{\frac{n^{2}-1}{12}} is the standard deviation of the first \mathrm{n} natural numbers.

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