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 the following frequency distribution is $50$ and $\Sigma f = 120$. Find the missing frequencies $f_1$ and $f_2$.

Class	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$
$f$	$17$	$f_1$	$32$	$f_2$	$19$

Let the mean and the variance of $20$ observations $x_{1}, x_{2}, \ldots, x_{20}$ be $15$ and $9$,respectively. For $\alpha \in R$,if the mean of $(x_{1}+\alpha)^{2}, (x_{2}+\alpha)^{2}, \ldots, (x_{20}+\alpha)^{2}$ is $178$,then the square of the maximum value of $\alpha$ is equal to $...........$

Statement $(I)$: The range of the ungrouped data does not change even if certain intermediate observations are removed.
Statement $(II)$: The value of the mean deviation of an ungrouped data about the median is always less than or equal to the value of the mean deviation computed about any other measure of central tendency.
Statement $(III)$: For a grouped data,range is approximated as the difference between the lower limit of the largest class and the upper limit of the smallest class.

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 $4$ observations are $\frac{7}{2}$ and $a$ respectively,then $(4a + x_{5})$ is equal to

The mean and the standard deviation of $10$ observations are $20$ and $2$ respectively. Each of these $10$ observations is multiplied by $p$ and then reduced by $q$,where $p \neq 0$ and $q \neq 0$. If the new mean and new standard deviation (s.d.) become half of the original values,then $q$ is equal to