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Question

# Let $\overline{)x}$ be the mean of ${x}_{1},{x}_{2},...,{x}_{n}$ and $\overline{)y}$ be the mean of ${y}_{1},{y}_{2},...,{y}_{n}$. If $\overline{)z}$ is the mean of ${x}_{1},{x}_{2},...,{x}_{n},{y}_{1},{y}_{2},...,{y}_{n},$ then $\overline{)z}=?$ (a) $\left(\overline{)x}+\overline{)y}\right)$ (b) $\frac{1}{2}\left(\overline{)x}+\overline{)y}\right)$ (c) $\frac{1}{n}\left(\overline{)x}+\overline{)y}\right)$ (d) $\frac{1}{2n}\left(\overline{)x}+\overline{)y}\right)$

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Solution

## (b) $\frac{1}{2}\left(\overline{)x}+\overline{)y}\right)$ $\overline{z}=\frac{\left({x}_{1}+{x}_{2}+...+{x}_{n}\right)+\left({y}_{1}+{y}_{2}+...+{y}_{n}\right)}{2n}$ $\mathrm{Given}:\phantom{\rule{0ex}{0ex}}\overline{x}=\frac{{x}_{1}+{x}_{2}+.....{x}_{n}}{n}\phantom{\rule{0ex}{0ex}}⇒{x}_{1}+{x}_{2}+......+{x}_{n}=n\overline{x}\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}\overline{y}=\frac{{y}_{1}+{y}_{2}+......+{y}_{n}}{n}\phantom{\rule{0ex}{0ex}}⇒{y}_{1}+{y}_{2}+......+{y}_{n}=n\overline{y}\phantom{\rule{0ex}{0ex}}\therefore \overline{z}=\frac{n\overline{x}+n\overline{y}}{2n}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(\overline{x}+\overline{y}\right)$

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