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 $a x_{1}, a x_{2}, a x_{3}, \ldots ., a x_{n}$ are $a \bar{x}$ and $a^{2} \sigma^{2},$ respectively, $(a \neq 0)$

The given n observations are $x _{1}, x _{2} \ldots x _{ n }$

Mean $=\bar{x}$

Variance $=\sigma^{2}$

$\therefore {\sigma ^2} = \frac{1}{n}\sum\limits_{i = 1}^n {{y_1}{{\left( {{x_i} - \bar x} \right)}^2}} $       ..........$(1)$

If each observation is multiplied by a and the new observations are $y_{i},$ then

$y_{1}=a x_{i}$ i.e., $x_{i}=\frac{1}{a} y_{1}$

$\bar y = \frac{1}{n}\sum\limits_{i = 1}^n {{y_1} = \frac{1}{n}\sum\limits_{i = 1}^n {a{x_1} = \frac{a}{n}} } \sum\limits_{i = 1}^n {{x_1} = ax} $         $\left( {\bar x = \frac{1}{n}\sum\limits_{i = 1}^n {{x_1}} } \right)$

Therefore, mean of the observations, $a x_{1}, a x_{2} \ldots a x_{n},$ is $a \bar{x}$

Substituting the values of $x_{i}$ and $\bar{x}$ in $(1),$ we obtain

${\sigma ^2} = \frac{1}{n}{\sum\limits_{i = 1}^n {\left( {\frac{1}{a}{y_1} - \frac{1}{a}\bar y} \right)} ^2}$

$ \Rightarrow {a^2}{\sigma ^2} = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{y_1} - \bar y} \right)}^2}} $

Thus, the variance of the observations, $a x_{1}, a x_{2} \ldots a x_{n},$ is $a^{2} \sigma^{2}$