The mean and variance of seven observations are $8$ and $16$ respectively. If five of the observations are $2, 4, 10, 12, 14$,then the product of the remaining two observations is:

The average marks of boys in a class is $40$ and that of girls is $45$. The average marks of both boys and girls combined is $42$. Then the percentage of boys in the class is (in $\%$)

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

$x_1, x_2, \ldots, x_n$ are $n$ observations with mean $\bar{x}$ and standard deviation $\sigma$. Match the items of List-$I$ with those of List-$II$:

List-$I$	List-$II$
$(a) \sum_{i=1}^n(x_i-\bar{x})$	$(i) \text{ Median}$
$(b) \text{ Variance } (\sigma^2)$	$(ii) \text{ Coefficient of variation}$
$(c) \text{ Mean deviation}$	$(iii) \text{ Zero}$
$(d) \text{ Measure used to find the homogeneity of given two series}$	$(iv) \text{ Mean of the absolute deviations from any measure of central tendency}$
	$(v) \text{ Mean of the squares of the deviations from mean}$

Let the mean and variance of $12$ observations be $\frac{9}{2}$ and $4$ respectively. Later on,it was observed that two observations were considered as $9$ and $10$ instead of $7$ and $14$ respectively. If the correct variance is $\frac{m}{n}$,where $m$ and $n$ are co-prime,then $m + n$ is equal to

If both mean and variance of $50$ observations $x_1, x_2, \ldots, x_{50}$ are equal to $16$ and $256$ respectively,then the mean of $(x_1-5)^2, (x_2-5)^2, \ldots, (x_{50}-5)^2$ is