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Question:

Let $\bar{x}$ be the mean of $x_{1}, x_{2}, \ldots, x_{n}$ and $\bar{y}$ be the mean of $y_{1}, y_{2}, \ldots, y_{n}$.

If $\bar{z}$ is the mean of $x_{1}, x_{2}, \ldots, x_{n}, y_{1}, y_{2}, \ldots, y_{n}$, then $\bar{z}=?$

(a) $(\bar{x}+\bar{y})$

(b) $\frac{1}{2}(\bar{x}+\bar{y})$

(c) $\frac{1}{n}(\bar{x}+\bar{y})$

(d) $\frac{1}{2 n}(\bar{x}+\bar{y})$

 

Solution:

(b) $\frac{1}{2}(\bar{x}+\bar{y})$

$\bar{z}=\frac{\left(x_{1}+x_{2}+\ldots+x_{n}\right)+\left(y_{1}+y_{2}+\ldots+y_{n}\right)}{2 n}$

Given :

$\bar{x}=\frac{x_{1}+x_{2}+\ldots \ldots x_{n}}{n}$

$\Rightarrow x_{1}+x_{2}+\ldots \ldots+x_{n}=n \bar{x}$

and

$\bar{y}=\frac{y_{1}+y_{2}+\ldots \ldots+y_{n}}{n}$

$\Rightarrow y_{1}+y_{2}+\ldots \ldots+y_{n}=n \bar{y}$

$\therefore \bar{z}=\frac{n \bar{x}+n \bar{y}}{2 n}$

$=\frac{1}{2}(\bar{x}+\bar{y})$

 

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