The mean of $5$ observations is $4.4$ and their variance is $8.24$. If three of the observations are $1, 2$ and $6$,find the other two observations.

If the mean deviation about the mean is $m$ and variance is $\sigma^2$ for the following data,then $m + \sigma^2 =$

$x$	$1, 3, 5, 7, 9$
$f$	$4, 24, 28, 16, 8$

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:

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 $100$ observations were calculated as $40$ and $5.1$ respectively. Later on,it was found that one of the observations was taken as $50$ in place of $40$. If the wrong entry is replaced by the correct one,then the sum of the squares of all the observations is:

The mean and standard deviation of $20$ observations are found to be $10$ and $2$ respectively. On rechecking,it was found that an observation $8$ was incorrect. If the incorrect observation $8$ is replaced by $12$,calculate the correct mean and standard deviation.