The average marks of boys in a class is $52$ and that of girls is $42$. The average marks of boys and girls combined is $50$. The percentage of boys in the class is:

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

The mean and standard deviation of a set of $n_{1}$ observations are $\bar{x}_{1}$ and $s_{1},$ respectively,while the mean and standard deviation of another set of $n_{2}$ observations are $\bar{x}_{2}$ and $s_{2},$ respectively. Show that the standard deviation of the combined set of $(n_{1}+n_{2})$ observations is given by $SD = \sqrt{\frac{n_{1}(s_{1})^{2}+n_{2}(s_{2})^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}(\bar{x}_{1}-\bar{x}_{2})^{2}}{(n_{1}+n_{2})^{2}}}$.

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 ...... .

The mean and standard deviation of $100$ observations were calculated as $40$ and $5.1$,respectively,by a student who took by mistake $50$ instead of $40$ for one observation. What are the correct mean and standard deviation?

The possible values of $x$ if the standard deviation $(SD)$ of the numbers $2, 3, 2x$,and $11$ is $3.5$ are equal to: