Question

Let $\overline{\mathrm{X}}$ be the mean of x₁,x₂,...,x_n, and $\overline{\mathrm{Y}}$ the mean of y₁,y₂,...,y_n, if $\overline{\mathrm{Z}}$ is the mean of x₁,x₂,...,x_n, y₁,y₂,...,y_n, then $\overline{\mathrm{Z}}$ is equal to

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

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)

Also, the mean of y₁, y₂,..., y_n is $\overline{Y}$.

$\therefore \overline{Y}=\frac{y_1+y_2+...+y_n}{n}$

$\Rightarrow y_1+y_2+...+y_n=n\overline{Y}$ .....(2)

Now, $x_1,x_2,...,x_n,y_1,y_2,...,y_n$ are 2n observations. The mean of these 2n observations is $\overline{Z}$.

$\therefore \overline{Z}=\frac{x_1+x_2+...+x_n+y_1+y_2+...+y_n}{2n}$

$\Rightarrow \overline{Z}=\frac{n\overline{X}+n\overline{Y}}{2n}$ [Using (1) and (2)]

$\Rightarrow \overline{Z}=\frac{n\left(\overline{X}+\overline{Y}\right)}{2n}$

$\Rightarrow \overline{Z}=\frac{\overline{X}+\overline{Y}}{2}$

Hence, the correct answer is option (b).