Identify the data set

The first $n$ even natural numbers are $2,4,6,\dots,2n.$

Formula for variance

Variance of observations $x_1,x_2,\dots,x_n$ is $\sigma^2=\frac{1}{n}\sum x_i^2-\left(\frac{1}{n}\sum x_i\right)^2.$

Use Statement-2 formulas

For the numbers $2,4,6,\dots,2n$, $\sum_{i=1}^n 2i=2\sum_{i=1}^n i=2\cdot \frac{n(n+1)}{2}=n(n+1).$

Also, \sum_{i=1}^n (2i)^2=4\sum_{i=1}^n i^2=4\cdot \frac{n(n+1)(2n+1)}{6}=rac{2n(n+1)(2n+1)}{3}.

Compute the mean

$\bar x=\frac{1}{n}\sum_{i=1}^n 2i=\frac{n(n+1)}{n}=n+1.$

Compute the variance

$\sigma^2=\frac{1}{n}\cdot \frac{2n(n+1)(2n+1)}{3}-(n+1)^2$ $=\frac{2(n+1)(2n+1)}{3}-(n+1)^2$ $=(n+1)\left(\frac{2(2n+1)}{3}-(n+1)\right)$ $=(n+1)\left(\frac{4n+2-3n-3}{3}\right)$ $=(n+1)\left(\frac{n-1}{3}\right)$ $=\frac{n^2-1}{3}.$

Check Statement-1

Statement-1 claims variance is $\frac{n^2-1}{4}.$

But we obtained $\frac{n^2-1}{3}.$

Hence, Statement-1 is false.

Check Statement-2

The standard formulas $\sum_{i=1}^n i=\frac{n(n+1)}{2}, \qquad \sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}$ are correct.

Hence, Statement-2 is true.

Choose the correct option

• Statement-1: False
• Statement-2: True

Therefore, the correct option is $\boxed{\text{D}}.$