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 given $n$ observations are $x_{1}, x_{2}, \ldots, x_{n}$.
Mean $= \bar{x}$.
Variance $= \sigma^{2} = \frac{1}{n} \sum_{i=1}^{n} (x_{i} - \bar{x})^{2}$.
Let the new observations be $y_{i} = a x_{i}$ for $i = 1, 2, \ldots, n$.
The new mean $\bar{y}$ is given by:
$\bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_{i} = \frac{1}{n} \sum_{i=1}^{n} (a x_{i}) = a \left( \frac{1}{n} \sum_{i=1}^{n} x_{i} \right) = a \bar{x}$.
The new variance $\sigma_{y}^{2}$ is given by:
$\sigma_{y}^{2} = \frac{1}{n} \sum_{i=1}^{n} (y_{i} - \bar{y})^{2} = \frac{1}{n} \sum_{i=1}^{n} (a x_{i} - a \bar{x})^{2}$.
$\sigma_{y}^{2} = \frac{1}{n} \sum_{i=1}^{n} a^{2} (x_{i} - \bar{x})^{2} = a^{2} \left( \frac{1}{n} \sum_{i=1}^{n} (x_{i} - \bar{x})^{2} \right)$.
Since $\sigma^{2} = \frac{1}{n} \sum_{i=1}^{n} (x_{i} - \bar{x})^{2}$,we have $\sigma_{y}^{2} = a^{2} \sigma^{2}$.
Thus,the mean is $a \bar{x}$ and the variance is $a^{2} \sigma^{2}$.