Variance of first $2 n$ natural numbers is

Variance of first $2 n$ natural numbers is
  1. $\frac{4 n^2+1}{12}$
  2. $\frac{(2 \mathrm{n}-1)^2}{12}$
  3. $\frac{\mathrm{n}^2}{3}-1$
  4. $\frac{4 n^2-1}{12}$

Solution

$\begin{aligned} \sigma^2=\frac{1}{2 n}\left[1^2+2^2+3^2+\ldots\right. & \left.+(2 n)^2\right] \\ & -\left(\frac{1+2+3+\ldots+2 n}{2 n}\right)^2\end{aligned}$ $\begin{aligned} & =\frac{1}{2 n}\left[\frac{2 n(2 n+1)(4 n+1)}{6}\right]-\left[\frac{1}{2 n} \times \frac{2 n(2 n+1)}{2}\right]^2 \\ & =\frac{(2 n+1)(4 n+1)}{6}-\left(\frac{2 n+1}{2}\right)^2 \\ & =\frac{2 n+1}{2}\left(\frac{4 n+1}{3}-\frac{2 n+1}{2}\right) \\ & =\frac{2 n+1}{2}\left(\frac{2 n-1}{6}\right) \\ & =\frac{4 n^2-1}{12}\end{aligned}$

Asked in: MHT CET 2023 (13 May Shift 2)

Practice more Statistics questions on Aicharya