If $\sum_{i=1}^{n}(x_{i}-a)=n$ and $\sum_{i=1}^{n}(x_{i}-a)^{2}=na$,where $n, a > 1$,then the standard deviation of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$ is

Let the observations $x_{i} (1 \leq i \leq 10)$ satisfy the equations $\sum_{i=1}^{10}(x_{i}-5)=10$ and $\sum_{i=1}^{10}(x_{i}-5)^{2}=40$. If $\mu$ and $\lambda$ are the mean and the variance of the observations $x_{1}-3, x_{2}-3, \dots, x_{10}-3$,then the ordered pair $(\mu, \lambda)$ is equal to:

The variance of the following frequency distribution is

Class Interval	Frequency
$0 - 6$	$10$
$6 - 12$	$8$
$12 - 18$	$6$
$18 - 24$	$4$
$24 - 30$	$2$

Find the variance of the following discrete frequency distribution:

Class Interval	$0-2$	$2-4$	$4-6$	$6-8$	$8-10$
Frequency $(f_i)$	$2$	$3$	$5$	$3$	$2$

Calculate the variance if $\Sigma x_i^2 = 18000$ and $\Sigma x_i = 960$,for $60$ observations.

If the variance of $6, 7, 8, 9, 10, 11$ is $\sigma^2$,then the variance of $12, 14, 16, 18, 20, 22$ is