Let the mean and variance of $8$ numbers $x, y, 10, 12, 6, 12, 4, 8$ be $9$ and $9.25$ respectively. If $x > y$,then $3x - 2y$ 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.

For two data sets,each of size $5$,the variances are given to be $4$ and $5$ and the corresponding means are given to be $2$ and $4$,respectively. The variance of the combined data set is

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	$(4) \ 145$
	$(5) \ 55$

What is the weighted mean of the first $n$ natural numbers when the weights are equal to the corresponding natural numbers?

If $\bar{x}_1$ and $\bar{x}_2$ are the means of two distributions such that $\bar{x}_1 < \bar{x}_2$ and $\bar{x}$ is the mean of the combined distribution,then