Mean of a certain number of observation is $\bar{x}$. If each observation is divided by $m(m \neq 0)$ and increased by $n$, then the mean of new observation is
(a) $\frac{\bar{x}}{\frac{m}{x}}+n$
(b) $\frac{\frac{m}{x}}{n}+m$
(c) $\frac{n}{x}+\frac{n}{m}$
(d) $\bar{x}+\frac{m}{n}$
Let $y_{1}, y_{2}, y_{3}, \ldots, y_{k}$ be $k$ observations.
Mean of the observations $=\bar{x}$
$\Rightarrow \frac{y_{1}+y_{2}+y_{3}+\ldots+y_{k}}{k}=\bar{x}$
$\Rightarrow y_{1}+y_{2}+y_{3}+\ldots+y_{k}=k \bar{x} \quad \ldots \ldots(1)$
If each observation is divided by m and increased by n, then the new observations are
$\frac{y_{1}}{m}+n, \frac{y_{2}}{m}+n, \frac{y_{3}}{m}+n, \ldots, \frac{y_{k}}{m}+n$
$\therefore$ Mean of new observations
$=\frac{\left(\frac{y_{1}}{m}+n\right)+\left(\frac{y_{2}}{m}+n\right)+\ldots+\left(\frac{y_{k}}{m}+n\right)}{k}$
$=\frac{\left(\frac{y_{1}}{m}+\frac{y_{2}}{m}+\ldots+\frac{y_{k}}{m}\right)+(n+n+\ldots+n)}{k}$
$=\frac{y_{1}+y_{2}+\ldots+y_{k}}{m k}+\frac{n k}{k}$
$=\frac{k \bar{x}}{m k}+\frac{n k}{k}$
$=\frac{\bar{x}}{m}+n$
Hence, the correct option is (a).