If each of the observations $x_{1}, x_{2}, \ldots, x_{n}$ is increased by $a$,where $a$ is a positive or negative number,show that the variance remains unchanged.

Let $\bar{x}$ be the mean of $x_{1}, x_{2}, \ldots, x_{n}$. The variance of the original observations is given by:
$\sigma_1^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2$
If $a$ is added to each observation,the new observations are $y_i = x_i + a$.
The new mean $\bar{y}$ is:
$\bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = \frac{1}{n} \sum_{i=1}^{n} (x_i + a) = \frac{1}{n} (\sum_{i=1}^{n} x_i + na) = \bar{x} + a$.
The variance of the new observations $\sigma_2^2$ is:
$\sigma_2^2 = \frac{1}{n} \sum_{i=1}^{n} (y_i - \bar{y})^2 = \frac{1}{n} \sum_{i=1}^{n} ((x_i + a) - (\bar{x} + a))^2$
$= \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 = \sigma_1^2$.
Thus,the variance remains unchanged.