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Let $\bar{x}$ be the mean of $x_1,\ x_2........., x_n$ and $\bar{y}$ the mean of $y_{1}, y_{2}, ... , y_{n}$ . If $\bar{z}$ is the mean of $x_1,\ x_2........., x_n$ , $y_{1}, y_{2}, ... , y_{n}$ then $\bar{z}$ is equal to

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

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

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

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

Answers (1)

Answer : B

$\bar{x}$ is the mean of $x_{1},x_{2},.....,x_{n}$ then

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

$\bar{y}$ is the mean of $y_{1},y_{2},......y_{n}$ then

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

$\bar{z}$ is the mean of $x_{1},x_{2},.......x_{n},y_{1},y_{2}............,y_{n}$

$\bar{z} = \frac{x_{1}+x_{2}+........+x_{n}+y_{1}+y_{2}+.....y_{n}}{2n}$

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

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

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

Hence, the value of $\bar{z}$ is $\frac{\bar{x}+\bar{y}}{2}$

Therefore option (B) is correct.

Posted by

infoexpert23

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