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

An online exam is attempted by $50$ candidates,out of which $20$ are boys. The average marks obtained by boys is $12$ with a variance of $2$. The variance of marks obtained by $30$ girls is also $2$. The average marks of all $50$ candidates is $15$. If $\mu$ is the average marks of girls and $\sigma^{2}$ is the variance of marks of $50$ candidates,then $\mu+\sigma^{2}$ is equal to ...... .

Consider the frequency distribution of the given numbers. If the mean is known to be $3$,then the value of $f$ is:

Value	$1$	$2$	$3$	$4$
Frequency	$5$	$4$	$6$	$f$

If the variance of the frequency distribution is $3$,then $\alpha$ is ......

$X_i$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
Frequency $f_i$	$3$	$6$	$16$	$\alpha$	$9$	$5$	$6$

The mean and standard deviation of $50$ observations are $15$ and $2$ respectively. It was found that one incorrect observation was taken such that the sum of the correct and incorrect observations is $70$. If the correct mean is $16$,then the correct variance is equal to

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 $\%$)