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Q.
The variance of first n natural numbers is :
Statistics
Solution:
We know that variance $\sigma^2=\frac{\sum x _1^2}{ n }-\left(\frac{\sum x _1}{ n }\right)^2$
First $n$ natural numbers are:
$\therefore \sum x _{ i }^2=\frac{ n ( n +1)(2 n +1)}{6}$
and $\sum x _{ i }=\frac{ n ( n +1)}{2}$
$\sigma^2=\frac{\sum x_{ i }^2}{n}-\left(\frac{\sum x_i}{n}\right)^2$
$=\frac{ n ( n +1)(2 n +1)}{6 n }-\left[\frac{ n ( n +1)}{2 n }\right]^2$
$=\frac{\left(2 n ^2+3 n +1\right)}{6}-\left(\frac{ n +1}{4}\right)^2$
$=\frac{2\left[2 n^2+3 n+1\right]-3\left(n^2+1+2 n\right)}{12}$
$=\frac{4 n ^2+6 n +2-3 n ^2-3-6 n }{12}=\frac{ n ^2-1}{12}$
