The mean and standard deviation of six observations are $8$ and $4,$ respectively. If each observation is multiplied by $3,$ find the new mean and new standard deviation of the resulting observations.

Given that $\bar{x}$ is the mean and $\sigma^{2}$ is the variance of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$,prove that the mean and variance of the observations $a x_{1}, a x_{2}, \ldots, a x_{n}$ are $a \bar{x}$ and $a^{2} \sigma^{2}$ respectively,where $a \neq 0$.

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 $S.D.$ of $15$ items is $6$. If each item is decreased or increased by $1$,then the standard deviation will be:

If the variance of the first $n$ natural numbers is $10$ and the variance of the first $m$ even natural numbers is $16$,then $n: m=$

The variance of the following continuous frequency distribution is
$\begin{array}{|l|c|c|c|c|}\hline \text{Class interval} & 0-4 & 4-8 & 8-12 & 12-16 \\ \hline \text{Frequency} & 2 & 3 & 2 & 1 \\ \hline\end{array}$