Let $\bar{x}$ be the mean of $x_1, x_2, x_3, \cdots, x_n$ and a be a nonzero real number.
Then, $\bar{x}=\frac{1}{n}\left(x_1+x_2+x_3+\cdots+x_n\right)$
Let $y_i=a x_i$ for each $i=1,2,3, \ldots, n$. Then,
$\begin{aligned} & \bar{y}=\frac{1}{n}\left(y_1+y_2+y_3+\cdots+y_n\right) \\ & =\frac{1}{n}\left(a x_1+a x_2+a x_3+\cdots+a x_n\right)=\text { a. } \frac{1}{n}\left(x_1+x_2+x_3+\cdots x_n\right)=a \bar{x}\end{aligned}$
Thus, $\bar{y}=a \bar{x}$
Now, the variance of new observations is given by
variance $(\mathrm{y})=\sigma_1^2$
$=\frac{1}{n} \sum_{i=1}^n\left(y_i-\bar{y}\right)^2$
$=\frac{1}{n} \sum_{i=1}^n\left(a x_i-a \bar{x}\right)^2\left[\because \quad y_i=a x_i\right.$ for each $i$ and $\left.\bar{y}=a \bar{x}\right]$
$=a^2 \cdot \frac{1}{n} \cdot \sum_{\mathrm{i}=1}^{\mathrm{n}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2$
$\begin{aligned} & =\mathrm{a}^2 .\{\text { variance }(\mathrm{x})\}=\mathrm{a}^2 \sigma^2 \\ & \therefore \text { new variance }=\mathrm{a}^2 \sigma^2 \\ & \text { Remark } \sigma_1=\sqrt{\mathrm{a}^2 \sigma^2}=|\mathrm{a}| \cdot \sigma\end{aligned}$