The average age of a group of men and women is $30$ years. If the average age of men is $32$ years and that of women is $27$ years,what is the percentage of women in the group?

Consider the following frequency distribution:

Value	$4$	$5$	$8$	$9$	$6$	$12$	$11$
Frequency	$5$	$f_1$	$f_2$	$2$	$1$	$1$	$3$

Suppose that the sum of the frequencies is $19$ and the median of this frequency distribution is $6$. For the given frequency distribution,let $\alpha$ denote the mean deviation about the mean,$\beta$ denote the mean deviation about the median,and $\sigma^2$ denote the variance. Match each entry in List-$I$ to the correct entry in List-$II$ and choose the correct option.

List-$I$	List-$II$
$(P) \ 7f_1+9f_2$ is equal to	$(1) \ 146$
$(Q) \ 19\alpha$ is equal to	$(2) \ 47$
$(R) \ 19\beta$ is equal to	$(3) \ 48$
$(S) \ 19\sigma^2$ is equal to	$(4) \ 145$
	$(5) \ 55$

The mean and standard deviation of $20$ observations are found to be $10$ and $2$,respectively. It was later discovered that one observation was mistakenly recorded as $8$ instead of $12$. The correct standard deviation is:

Suppose that the mean and median of the non-negative numbers $21, 8, 17, a, 51, 103, b, 13, 67$ $(a > b)$ are $40$ and $21$,respectively. If the mean deviation about the median is $26$,then $2a$ is equal to:

The mean age of $25$ teachers in a school is $40 \text{ years}$. $A$ teacher retires at the age of $60 \text{ years}$ and a new teacher is appointed in his place. If now the mean age of the teachers in this school is $39 \text{ years}$,then the age (in years) of the newly appointed teacher is

There are $60$ students in a class. The following is the frequency distribution of the marks obtained by the students in a test:
$\begin{array}{|l|l|l|l|l|l|l|} \hline \text{Marks} & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \text{Frequency} & x-2 & x & x^2 & (x+1)^2 & 2x & x+1 \\ \hline \end{array}$
where $x$ is a positive integer. Determine the mean and standard deviation of the marks.