The mean and variance of a set of $6$ terms are $11$ and $24$ respectively,and the mean and variance of another set of $3$ terms are $14$ and $36$ respectively. Then,the variance of all $9$ terms is equal to:

The mean of $100$ items is $49$. It was discovered that three items which should have been $60, 70, 80$ were wrongly read as $40, 20, 50$ respectively. The correct mean is

Let $a, b, c \in N$ and $a < b < c$. Let the mean,the mean deviation about the mean,and the variance of the $5$ observations $9, 25, a, b, c$ be $18, 4$,and $\frac{136}{5}$,respectively. Then $2a + b - c$ is equal to:

If the algebraic sum of deviations of $20$ observations from $30$ is $20$,then the mean of the observations is:

If the variance of the frequency distribution is $3$,then $\alpha$ is ......

$X_i$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
Frequency $f_i$	$3$	$6$	$16$	$\alpha$	$9$	$5$	$6$

The mean income of a group of individuals is $400$ and the mean income of another group of individuals is $480$. If the mean income of all the individuals in both groups combined is $430$,find the ratio of the number of individuals in the two groups.