If the coefficient of variation and standard deviation are $ 60 $ and $ 21 $ respectively,the arithmetic mean of the distribution is:

The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results:
Number of observations $= 25$,mean $= 18.2 \text{ seconds}$,standard deviation $= 3.25 \text{ s}$.
Further,another set of $15$ observations $x_{1}, x_{2}, \ldots, x_{15}$,also in seconds,is now available and we have $\sum_{i=1}^{15} x_{i} = 279$ and $\sum_{i=1}^{15} x_{i}^{2} = 5524$. Calculate the standard deviation based on all $40$ observations.

The sum of squares of deviations for $10$ observations taken from mean $50$ is $250$. The coefficient of variation is.....$\%$

If $\sum_{i=1}^{10} (x_i - 15) = 12$ and $\sum_{i=1}^{10} (x_i - 15)^2 = 18$,find the standard deviation of the observations $x_1, x_2, \dots, x_{10}$.

If $v$ is the variance and $\sigma$ is the standard deviation,then

If the variance of $6, 7, 8, 9, 10, 11$ is $\sigma^2$,then the variance of $12, 14, 16, 18, 20, 22$ is