Given that $\bar{X}$ is the mean and ${\sigma }^2$ is the variance of $n$ observations $X_1,X_2...X_n$. Prove that the mean and variance of the observations $a{X}_1,a{X}_2,a{X}_3....a{X}_n$ are $\overline{ax}$ and $a^2{\sigma }^2$ respectively $(a\ne 0)$.

The given  $n$ observation are $x_1,x_2..x_n$
 Mean = $\bar{x}$ 
variance = ${\sigma }^2$
$\therefore {\sigma }^2=\frac{1}{n}\sum _{i=l}^n{y}_i{\left(x_i-\bar{x}\right)}^2$                    ....(i)
If each observation is multiplied by a and the new observation are $y_i$ then 
$y_i=a{x}_i,i.e,x_i=\frac{1}{a}{y}_i$
$\therefore \bar{y}=\frac{1}{n}\sum _{i=l}^n{y}_i=\frac{1}{n}\sum _{i=l}^{n}a{x}_i=\frac{a}{n}\sum _{i=l}^n{x}_i=\overline{ax},\left(\because \bar{x}=\frac{1}{n}\sum _{i=1}^n{x}_i\right)$
Therefore mean of the observation, $a{x}_1,a{x}_2....a{x}_n$ is $\overline{ax}$ 
Substituting the values of $x_i$ and $\bar{x}$ in (1) we obtain 
${\sigma }^2=\frac{1}{2}\sum _{i=l}^n{\left(\frac{1}{a}{y}_i-\frac{1}{a}\bar{y}\right)}^2$
$\Rightarrow a^2{\sigma }^2=\frac{1}{n}\sum _{i=l}^n{\left(y_i-\bar{y}\right)}^2$
Thus the variance of the observation $a{x}_1,a{x}_2...a{x}_n$ is $a^2{\sigma }^2$