Question

State the following statement is True or False

The variance of first $n$ even natural numbers is $\frac{n^2-1}{4}$

• A. True
• B. False

Answer 1

The correct option is B False

Mean $=\frac{{\sum }_{k=1}^{n}2k}{n}$

$\overline{x}=$ Mean $=n+1$

Variance $=\frac{\sum (x-\overline{x}{)}^2}{n}$

$\frac{\sum (x^2-2x\overline{x}+x^{-2})}{n}$

$=\frac{\frac{4\left(n\right)(n+1)(2n+1)}{6}-\frac{4n(n+1{)}^2}{2}=n(n+1{)}^2}{n}$

$=\frac{2}{3}(n+1)(2n+1)-(n+1{)}^2$

Variance $=\frac{n^2-1}{3}$