If the variance of the numbers $9, 15, 21, \ldots, (6n+3)$ is $P$,then the variance of the first $n$ even numbers is

Statement-$1$: The variance of the first $n$ even natural numbers is $\frac{n^2 - 1}{3}$.
Statement-$2$: The sum of the first $n$ odd natural numbers is $n^2$ and the sum of the squares of the first $n$ odd natural numbers is $\frac{n(4n^2 - 1)}{3}$.

The standard deviation of the following distribution is:

Class interval	$0-10$	$10-20$	$20-30$	$30-40$
Frequency	$1$	$3$	$4$	$2$

The mean and standard deviation of $100$ items are $50$ and $4$ respectively. The sum of the squares of all the items is:

The mean and standard deviation of $100$ items are $50$ and $4$, respectively. Then the sum of the squares of all the items is:

Consider $10$ observations $x_1, x_2, \ldots, x_{10}$ such that $\sum_{i=1}^{10}(x_i-\alpha)=2$ and $\sum_{i=1}^{10}(x_i-\beta)^2=40$,where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. The value of $\frac{\beta}{\alpha}$ is equal to :