The variance of the following frequency distribution is:

Class Interval	$0$-$4$	$4$-$8$	$8$-$12$	$12$-$16$	$16$-$20$
Frequency	$2$	$4$	$6$	$3$	$1$

If both the mean and the standard deviation of $50$ observations $x_1, x_2, \dots, x_{50}$ are equal to $16$,then the mean of $(x_1 - 4)^2, (x_2 - 4)^2, \dots, (x_{50} - 4)^2$ is

The frequency distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline X & A & 2 A & 3 A & 4 A & 5 A & 6 A \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$
where $A$ is a positive integer,has a variance of $160$. Determine the value of $A$.

If each observation of a raw data whose variance is $\sigma^2$ is multiplied by $\lambda$,then the variance of the new set is

Let $a, b, c, d$ and $e$ be the observations with mean $m$ and standard deviation $S$. The standard deviation of the observations $a+k, b+k, c+k, d+k$ and $e+k$ 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: