If the mean of observation $x_{1}, x_{2}, \ldots, x_{n}$ is $\bar{x}$, then the mean of $x_{1}+a, x_{2}+a_{1} \ldots \ldots, x_{n}+a$ is
(a) $a \bar{x}$
(b) $\bar{x}-a$
(c) $\bar{x}+a$
(d) $\frac{\bar{x}}{a}$
The mean of $x_{1}, x_{2}, \ldots, x_{n}$ is $\bar{x}$.
$\therefore \frac{x_{1}+x_{2}+x_{3}+\ldots+x_{n}}{n}=\bar{x}$
$\Rightarrow x_{1}+x_{2}+x_{3}+\ldots+x_{n}=n \bar{x}$
Mean of $x_{1}+a_{1} x_{2}+a, \ldots, x_{n}+a$
$=\frac{\left(x_{1}+a\right)+\left(x_{2}+a\right)+\left(x_{3}+a\right)+\ldots+\left(x_{n}+a\right)}{n}$
$=\frac{\left(x_{1}+x_{2}+x_{3}+\ldots+x_{n}\right)+(a+a+a+\ldots+a)}{n}$
$=\frac{n \bar{x}+n a}{n}$
$=\bar{x}+a$
Hence, the correct option is (c).
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