The mean and variance of a set of $6$ terms are $11$ and $24$ respectively,and the mean and variance of another set of $3$ terms are $14$ and $36$ respectively. Then,the variance of all $9$ terms is equal to:

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 $SD = \sqrt{\frac{n_{1}(s_{1})^{2}+n_{2}(s_{2})^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}(\bar{x}_{1}-\bar{x}_{2})^{2}}{(n_{1}+n_{2})^{2}}}$.

Let the median and the mean deviation about the median of $7$ observations $170, 125, 230, 190, 210, a, b$ be $170$ and $\frac{205}{7}$ respectively. Then the mean deviation about the mean of these $7$ observations is:

The mean of $5$ observations is $4.4$ and the variance is $8.24$. If three of the five observations are $1, 2,$ and $6$,then the values of the other two observations are:

The mean and variance of $8$ observations are $10$ and $13.5,$ respectively. If $6$ of these observations are $5, 7, 10, 12, 14, 15,$ then the absolute difference of the remaining two observations is

The mean of $n$ values of a distribution is $\bar{x}$. If the first value is increased by $1$,the second value by $2$,and so on,what will be the mean of the new values?