In a distribution of $10$ observations,the sum of the observations is $60$ and the sum of their squares is $1000$. Then,the variance is:

The $S.D.$ of the first $n$ natural numbers is

For $10$ observations $x_1, x_2, ..., x_{10}$, if $\sum_{i=1}^{10} (x_i + 2)^2 = 180$ and $\sum_{i=1}^{10} (x_i - 1)^2 = 90$, then their standard deviation is:

If $\sum_{i=1}^{18} (x_i - 8) = 9$ and $\sum_{i=1}^{18} (x_i - 8)^2 = 45$,find the standard deviation of $x_1, x_2, \dots, x_{18}$.

If both the mean and the standard deviation of $50$ observations $x_1, x_2, \dots, x_{50}$ are equal to $16$,then the mean of $(x_1 - 4)^2, (x_2 - 4)^2, \dots, (x_{50} - 4)^2$ is

The variance of the numbers $8, 21, 34, 47, \ldots, 320$ is . . . . . . .