Question

If variance of the first $n$ natural numbers is $10,$ then the value of $n$ is

Answer 1

Variance, ${\sigma }^2=\frac{\sum x_i^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2$
For first $n$ natural numbers,
$\frac{\sum x_i^2}{n}=\frac{1^2+2^2+3^2+\cdots +n^2}{n}\Rightarrow \frac{\sum x_i^2}{n}=\frac{n(n+1)(2n+1)}{6n}\Rightarrow \frac{\sum x_i^2}{n}=\frac{(n+1)(2n+1)}{6}$

$\frac{\sum x_i}{n}=\frac{1+2+3+\cdots +n}{n}\Rightarrow \frac{\sum x_i}{n}=\frac{n(n+1)}{2n}=\frac{n+1}{2}$
Now, ${\sigma }^2=\frac{(n+1)(2n+1)}{6}-\frac{(n+1{)}^2}{4}$
$\Rightarrow {\sigma }^2=\frac{(n+1)(2n-2)}{24}\Rightarrow \frac{n^2-1}{12}=10\therefore n=11$