If $\bar{x}_1$ and $\bar{x}_2$ are the means of two distributions such that $\bar{x}_1 < \bar{x}_2$ and $\bar{x}$ is the mean of the combined distribution,then

Let the mean and the variance of $20$ observations $x_{1}, x_{2}, \ldots, x_{20}$ be $15$ and $9$,respectively. For $\alpha \in R$,if the mean of $(x_{1}+\alpha)^{2}, (x_{2}+\alpha)^{2}, \ldots, (x_{20}+\alpha)^{2}$ is $178$,then the square of the maximum value of $\alpha$ is equal to $...........$

The mean and variance of eight observations are $9$ and $9.25,$ respectively. If six of the observations are $6, 7, 10, 12, 12,$ and $13,$ find the remaining two observations.

The mean and the standard deviation of $10$ observations are $20$ and $2$ respectively. Each of these $10$ observations is multiplied by $p$ and then reduced by $q$,where $p \neq 0$ and $q \neq 0$. If the new mean and new standard deviation (s.d.) become half of the original values,then $q$ is equal to

The mean and variance of $5$ observations are $5$ and $8$ respectively. If $3$ observations are $1, 3, 5$,then the sum of cubes of the remaining two observations is

If the mean and variance of eight numbers $3, 7, 9, 12, 13, 20, x$,and $y$ are $10$ and $25$ respectively,then $x \cdot y$ is equal to