The mean and variance of six observations are $6$ and $12$ respectively. If each observation is multiplied by $3$,then the new variance of the resulting observations is:

The variance of the discrete data $3, 4, 5, 6, 7, 8, 10, 13$ is

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

The variance of the first ten multiples of $3$ is

The mean of $5$ observations is $7$. If four of these observations are $6, 7, 8, 10$ and one is missing,then the variance of all the five observations 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.