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Question

# Let $\overline{\mathrm{X}}$ be the mean of x1,x2,...,xn, and $\overline{\mathrm{Y}}$ the mean of y1,y2,...,yn, if $\overline{\mathrm{Z}}$ is the mean of x1,x2,...,xn, y1,y2,...,yn, then $\overline{\mathrm{Z}}$ is equal to $\left(a\right)\overline{\mathrm{X}}+\overline{\mathrm{Y}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(b\right)\frac{\overline{\mathrm{X}}+\overline{\mathrm{Y}}}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{c}\right)\frac{\overline{\mathrm{X}}+\overline{\mathrm{Y}}}{n}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{d}\right)\frac{\overline{\mathrm{X}}+\overline{\mathrm{Y}}}{2n}$

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Solution

## It is given that, the mean of x1, x2,..., xn is $\overline{X}$. $\therefore \overline{X}=\frac{{x}_{1}+{x}_{2}+...+{x}_{n}}{n}$ $\left(\mathrm{Mean}=\frac{\mathrm{Sum}\mathrm{of}\mathrm{observations}}{\mathrm{Number}\mathrm{of}\mathrm{observations}}\right)$ $⇒{x}_{1}+{x}_{2}+...+{x}_{n}=n\overline{X}$ .....(1) Also, the mean of y1, y2,..., yn is $\overline{Y}$. $\therefore \overline{Y}=\frac{{y}_{1}+{y}_{2}+...+{y}_{n}}{n}$ $⇒{y}_{1}+{y}_{2}+...+{y}_{n}=n\overline{)Y}$ .....(2) Now, ${x}_{1},{x}_{2},...,{x}_{n},{y}_{1},{y}_{2},...,{y}_{n}$ are 2n observations. The mean of these 2n observations is $\overline{Z}$. $\therefore \overline{)Z}=\frac{{x}_{1}+{x}_{2}+...+{x}_{n}+{y}_{1}+{y}_{2}+...+{y}_{n}}{2n}$ $⇒\overline{)Z}=\frac{n\overline{)X}+n\overline{)Y}}{2n}$ [Using (1) and (2)] $⇒\overline{)Z}=\frac{n\left(\overline{)X}+\overline{)Y}\right)}{2n}$ $⇒\overline{)Z}=\frac{\overline{)X}+\overline{)Y}}{2}$ Hence, the correct answer is option (b).

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