Find the standard deviation of the first n natural numbers. - Vidyadip

Question

Find the standard deviation of the first n natural numbers.

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Answer

x_i	1	2	3	4	5	-	-	n
x_i²	1	4	9	16	25	-	-	n²

Solution:
$\sum\text{x}_\text{i}=1+2+3+4+5+\ ....\ +\text{n}=\frac{\text{n}(\text{n}+1)}{2}$
$\sum\text{x}_\text{i}^2=1^2+2^2+3^2+\ ....\ +\text{n}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$\therefore\ \text{S.D.}(\sigma)=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{n}}\Big)^2}$
$=\sqrt{\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6\text{n}}-\frac{\text{n}^2(\text{n}+1)}{4\text{n}^2}}$
$=\sqrt{\frac{(\text{n}+1)(2\text{n}+1)}{6}-\frac{(\text{n+1}^2)}{4}}$
$=\sqrt{\frac{2\text{n}^2+3\text{n}+1}{6}-\frac{\text{n}^2+2\text{n}+1}{4}}$
$=\sqrt{\frac{4\text{n}^2+6\text{n}+2-3\text{n}^2-6\text{n}-3}{12}}$
$=\sqrt{\frac{\text{n}^2-1}{12}}$
Hence, the required SD $=\sqrt{\frac{\text{n}^2-1}{12}}$

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