Let $\bar{x}$ be the mean of $x_{1}, x_{2}, \ldots ., x_{n} .$ Then the variance is given by

$\sigma _1^2 = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} – \bar x} \right)}^2}} $

If $'a$ is added to each observation, the new observations will be

$y_{i}=x_{i}+a$        …….$(1)$

Let the mean of the new observations be $\bar{y} .$ Then

$\bar y = \frac{1}{n}\sum\limits_{i = 1}^n {{y_i} = \frac{1}{n}} \sum\limits_{i = 1}^n {\left( {{x_i} – a} \right)} $

$ = \frac{1}{n}\left[ {\sum\limits_{i = 1}^n {{x_i}} \sum\limits_{i = 1}^n a } \right] = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i} + \frac{{na}}{n} = } \bar x + a$

i.e.        $\bar{y}=\bar{x}+a$           ……….$(2)$

Thus, the variance of the new observations

$\sigma _2^2 = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{y_i} – \bar y} \right)}^2}}  = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} + a – \bar x – a} \right)}^2}} $         [ Using $(1)$ and $(2)$ ]

$ = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} + \bar x} \right)}^2}}  = \sigma _1^2$

Thus, the variance of the new observations is same as that of the original observations.