The mortality in a town during $4$ quarters of a year due to various causes is given in the bar graph below. Based on this data,calculate the percentage increase in mortality in the third quarter compared to the second quarter.

Let $x_1, x_2, \ldots, x_{11}$ be the observations satisfying $\sum_{i=1}^{11}(x_i-4)=22$ and $\sum_{i=1}^{11}(x_i-4)^2=154$. If the mean and variance of the observations are $\alpha$ and $\beta$,then the quadratic equation having the roots $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is

The mean of $6$ distinct observations is $6.5$ and their variance is $10.25$. If $4$ out of $6$ observations are $2, 4, 5$ and $7$,then the remaining two observations are:

The mean and variance of $7$ observations are $8$ and $16$ respectively. If five of the observations are $2, 4, 10, 12, 14$,find the remaining two observations.

The mean of a series $x_1, x_2, \dots, x_n$ is $\bar{X}$. If $x_2$ is replaced by $\lambda$,what will be the new mean?

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