Question

If $\overline{X}$ is the mean of x₁,x₂,.......,x_n , then for $a\ne 0$, the mean of ax₁,ax₂,......,ax_n, $\frac{x_1}{a},\frac{x_2}{a},...,\frac{x_n}{n}\mathrm{is}$

$$\begin{gathered} \left(\mathrm{a}\right)\left(a+\frac{1}{a}\right)\overline{\mathrm{X}} \\ \left(\mathrm{b}\right)\frac{1}{2}\left(a+\frac{1}{a}\right)\overline{\mathrm{X}} \\ \left(\mathrm{c}\right)\left(a+\frac{1}{n}\right)\frac{\overline{\mathrm{X}}}{n} \\ \left(\mathrm{d}\right)\left(a+\frac{1}{a}\right)\frac{\overline{\mathrm{X}}}{2n} \end{gathered}$$

Answer 1

It is given that, the mean of x₁, x₂,..., x_n is $\overline{X}$.

$\therefore \overline{X}=\frac{x_1+x_2+...+x_n}{n}$ $\left(\mathrm{Mean}=\frac{\mathrm{Sum}\mathrm{of}\mathrm{observations}}{\mathrm{Number}\mathrm{of}\mathrm{observations}}\right)$

$\Rightarrow x_1+x_2+...+x_n=n\overline{X}$ .....(1)

Now, $a{x}_1,a{x}_2,...,a{x}_n,\frac{x_1}{a},\frac{x_2}{a},...,\frac{x_n}{a}$ are 2n observations.

∴ Mean of these observations

$=\frac{a{x}_1+a{x}_2+...+a{x}_n+{\displaystyle \frac{x_1}{a}}+{\displaystyle \frac{x_2}{a}}+...+{\displaystyle \frac{x_n}{a}}}{2n}$

$=\frac{a\left(x_1+x_2+...+x_n\right)+{\displaystyle \frac{1}{a}}\left(x_1+x_2+...+x_n\right)}{2n}$

$=\frac{\left(a+{\displaystyle \frac{1}{a}}\right)\left(x_1+x_2+...+x_n\right)}{2n}$

$=\frac{\left(a+{\displaystyle \frac{1}{a}}\right)\times n\overline{X}}{2n}$ [Using (1)]

$=\frac{1}{2}\left(a+\frac{1}{a}\right)\overline{X}$

Thus, the mean of $a{x}_1,a{x}_2,...,a{x}_n,\frac{x_1}{a},\frac{x_2}{a},...,\frac{x_n}{a}$ is $\frac{1}{2}\left(a+\frac{1}{a}\right)\overline{X}$.

Hence, the correct answer is option (b).