Let $\mathrm{X}$ be a random variable with distribution.

$\mathrm{x}$	$-2$	$-1$	$3$	$4$	$6$
$\mathrm{P}(\mathrm{X}=\mathrm{x})$	$\frac{1}{5}$	$\mathrm{a}$	$\frac{1}{3}$	$\frac{1}{5}$	$\mathrm{~b}$

If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :

Let the mean and the variance of $5$ observations $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be $\frac{24}{5}$ and $\frac{194}{25}$ respectively. If the mean and variance of the first $4$ observation are $\frac{7}{2}$ and $a$ respectively, then $\left(4 a+x_{5}\right)$ is equal to

In a series of $2n$ observations, half of them equal to $a$ and remaining half equal to $-a$. If the standard deviation of the observations is $2$, then $|a|$ equals

If each of given $n$ observations is multiplied by a certain positive number $'k'$, then for new set of observations -

Determine the mean and standard deviation for the following distribution:

$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|} \hline \text { Marks } & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline \text { Frequency } & 1 & 6 & 6 & 8 & 8 & 2 & 2 & 3 & 0 & 2 & 1 & 0 & 0 & 0 & 1 \\ \hline \end{array}$

Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$ be in $A.P.$ and $a_1+a_3=10$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^2$, then $8 \sigma^2$ is equal to