Suppose a class has $7$ students. The average marks of these students in the mathematics examination is $62$,and their variance is $20$. $A$ student fails in the examination if they get less than $50$ marks. In the worst-case scenario,what is the maximum number of students who can fail?

An analysis of monthly wages paid to the workers of two jute mills $A$ and $B$ gives the following data:

Metric	Mill-$A$	Mill-$B$
No. of workers	$500$	$600$
Average daily wage (in rupees)	$186$	$175$
Variance of distribution of wages	$81$	$100$

Then:

If $M$ and $\sigma^2$ represent respectively the mean deviation from the mean and the variance for the data $1, 3, 5, 7, 11, 13, 17, 19, 23$,then $3(\sigma^2 - M) = $

Let ${x_1}, {x_2}, ..., {x_n}$ be $n$ observations such that $\sum x_i^2 = 400$ and $\sum x_i = 80$. Then a possible value of $n$ among the following is:

If the mean of two samples of sizes $20$ and $30$ are $25$ and $10$ respectively,and their variances are $9$ and $16$ respectively,then their combined variance is:

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 $SD = \sqrt{\frac{n_{1}(s_{1})^{2}+n_{2}(s_{2})^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}(\bar{x}_{1}-\bar{x}_{2})^{2}}{(n_{1}+n_{2})^{2}}}$.