The mean of the numbers $a, b, 8, 5, 10$ is $6$ and their variance is $6.8$. If $M$ is the mean deviation of the numbers about the mean,then $25M$ is equal to

The number of values of $a \in N$ such that the variance of $3, 7, 12, a, 43-a$ is a natural number is (Mean $= 13$).

The mean of $100$ observations is $45$. It was later found that two observations $19$ and $31$ were incorrectly recorded as $91$ and $13$. The correct mean is...

The frequency distribution of the daily household expenditure of some families in a residential area is given below. If the mode of the distribution is $140$,what is the value of $b$?

Daily Expenditure (	$0-50$	$50-100$	$100-150$	$150-200$	$200-250$
Number of Families $(f)$	$24$	$33$	$37$	$b$	$25$

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

Let $\bar{x}, M$ and $\sigma^2$ be respectively the mean,mode and variance of $n$ observations $x_1, x_2, ..., x_n$ and $d_i = -x_i - a, i = 1, 2, ..., n$,where $a$ is any number. Statement $I$: Variance of $d_1, d_2, ..., d_n$ is $\sigma^2$. Statement $II$: Mean and mode of $d_1, d_2, ..., d_n$ are $-\bar{x} - a$ and $-M - a$,respectively.