Variance of first $n$ natural numbers is $\qquad$ .

The standard deviation and mean of five observations are $0$ and $9$ respectively. If one of the observations is changed such that the mean of the new set of five observations becomes $10$,then their standard deviation is

Let $a_1, a_2, \ldots, a_{10}$ be $10$ observations such that $\sum_{k=1}^{10} a_k = 50$ and $\sum_{k < j} a_k a_j = 1100$. Then the standard deviation of $a_1, a_2, \ldots, a_{10}$ is equal to:

Let $x_1, x_2, \ldots, x_{10}$ be ten observations such that $\sum_{i=1}^{10}(x_i-2)=30$,$\sum_{i=1}^{10}(x_i-\beta)^2=98$,$\beta > 2$ and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of $2(x_1-1)+4\beta, 2(x_2-1)+4\beta, \ldots, 2(x_{10}-1)+4\beta$,then $\frac{\beta\mu}{\sigma^2}$ is equal to:

$\text{Assertion (A):}$ Variance of $4x_1, 4x_2, \ldots, 4x_n$ is $16$ times the variance of $x_1, x_2, \ldots, x_n$. $\text{Reason (R):}$ If $y = ax + b$,then variance of $y$ is $a(\text{variance of } x) + b$. The correct option among the following is

For the following frequency distribution,the variance is approximately equal to

Class Interval	$0$-$5$	$5$-$10$	$10$-$15$	$15$-$20$	$20$-$25$
Frequency	$4$	$1$	$10$	$3$	$2$