Let ${x_1}, {x_2}, ..., {x_n}$ be $n$ observations such that $\sum x_i^2 = 400$ and $\sum x_i = 80$. Then a possible value of $n$ among the following is:

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

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

The coefficients of variation of two distributions are $60$ and $70$. The standard deviations are $21$ and $16$ respectively. Find their means.

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:

If the coefficient of variation is $ 60 $ and the standard deviation is $ 24 $,then the arithmetic mean is: