$A$ market with $3900$ operating firms has the following distribution for firms arranged according to various income groups of workers. If a histogram for the above distribution is constructed,the highest bar in the histogram would correspond to the class:

Income group	No. of firms
$150-300$	$300$
$300-500$	$500$
$500-800$	$900$
$800-1200$	$1000$
$1200-1800$	$1200$

The mean and standard deviation of a set of $n_{1}$ observations are $\bar{x}_{1}$ and $s_{1},$ respectively,while the mean and standard deviation of another set of $n_{2}$ observations are $\bar{x}_{2}$ and $s_{2},$ respectively. Show that the standard deviation of the combined set of $(n_{1}+n_{2})$ observations is given by $SD = \sqrt{\frac{n_{1}(s_{1})^{2}+n_{2}(s_{2})^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}(\bar{x}_{1}-\bar{x}_{2})^{2}}{(n_{1}+n_{2})^{2}}}$.

If $\alpha$ and $\beta$ are respectively the mean deviation about the mean and the variance of the first five prime numbers,then the ordered pair $(\alpha, \beta)$ is:

Two data sets each of size $5$ are given. If the variances are $4$ and $5$ and their corresponding means are $2$ and $4$ respectively,then what is the variance of the combined data set?

Let the mean and variance of the frequency distribution be $6$ and $6.8$ respectively.

$x$	$2$	$6$	$8$	$9$
$f$	$4$	$4$	$\alpha$	$\beta$

If $x_{3}$ is changed from $8$ to $7$,then the mean for the new data will be:

Let sets $A$ and $B$ have $5$ elements each. Let the mean of the elements in sets $A$ and $B$ be $5$ and $8$ respectively,and the variance of the elements in sets $A$ and $B$ be $12$ and $20$ respectively. $A$ new set $C$ of $10$ elements is formed by subtracting $3$ from each element of $A$ and adding $2$ to each element of $B$. Then the sum of the mean and variance of the elements of $C$ is $.......$.