Mathematics - Statistics Question with Solution | TestHub

Statement-2 is true Statement-1: Sum of $n$ even natural numbers $=n(n+1)$ $$\begin{array}{rl}& \text{ Mean }(\bar{x})=\frac{n(n+1)}{n}=n+1\\ & \text{ Variance }=\left[\frac{1}{n}\sum {\left(x_1\right)}^2\right]-(\bar{x}{)}^2=\frac{1}{n}\left[2^2+4^2+\dots .+(2n{)}^2\right]-(n+1{)}^2\\ & =\frac{1}{n}{2}^2\left[1^2+2^2+\dots .+n^2\right]-(n+1{)}^2=\frac{4}{n}\frac{n(n+1)(2n+1)}{6}-(n+1{)}^2\\ & =\frac{(n+1)[2(2n+1)-3(n+1)]}{3}=\frac{(n+1)[4n+2-3n-3]}{3}=\frac{(n+1)(n-1)}{3}=\frac{n^2-1}{3}\end{array}$$ $\therefore$ Statement 1 is false.