If the standard deviation of first $n$ natural numbers is 2 , then the value of $n$ is

Question

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { S.D. }=\sqrt{\frac{\sum x_i^2}{n}-(\bar{x})^2} \ & \Rightarrow 2=\sqrt{\frac{\sum n^2}{n}-\left(\frac{\sum n}{n}ight)^2} \ & \Rightarrow 4=\frac{n(n+1)(2 n+1)}{6 n}-\left(\frac{n(n+1)}{2 n}ight)^2 \ & \Rightarrow 4=\frac{n+1}{2}\left\{\frac{2 n+1}{3}-\frac{n+1}{2}ight\} \ & \Rightarrow 4=\frac{n+1}{2}\left(\frac{n-1}{6}ight) \ & \Rightarrow n^2=49 \Rightarrow n=7\end{aligned}$

If the standard deviation of first $n$ natural numbers is 2 , then the value of $n$ is

Solution