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If $\bar{x}_{1},\bar{x}_{2},\bar{x}_{3},........\bar{x}_{n},$ are the means of n groups with $n_{1},n_{2},......n_{n}$ number of observations respectively, then the mean $\bar{x}$ of all the groups taken together is given by :

$(A)$ $\sum_{i=1}^{n}n_{i}\bar{x}_{i}$

$(B)$ $\frac{\sum_{i=1}^{n}n_{i}\bar{x}_{i}}{n^{2}}$

$(C)$ $\frac{\sum_{i=1}^{n}n_{i}\bar{x}_{i}}{\sum_{i=1}^{n}n_{i}}$

$(D)$ $\frac{\sum_{i=1}^{n}n_{i}\bar{x}_{i}}{2n}$

Answers (1)

Answer : C

$\text {Mean is defined as}\; \bar{x}=\frac{\text {sum of observations}}{\text {Number of observations}}$

$\Rightarrow$ Sum of observations = (mean) (Number of observations).

Given : $\bar{x}_{1},\bar{x}_{2},\bar{x}_{3},........\bar{x}_{n}$ are the means of n groups with $n_{1},n_{2},......n_{n}$ number of observations respectively.

$\sum x_{1}=\bar{x}_{1}n_{1}$

$\sum x_{2}=\bar{x}_{2}n_{2}$

$\sum x_{n}=\bar{x}_{n}n_{n}$

Now all the groups are taken together

Sum of all the observations:

$\sum x_{1}+\sum x_{2}+.......+\sum x_{n}=\bar{x}_{1}n_{1}+\bar{x}_{2}n_{2}+....+\bar{x}_{n}n_{n}$

Mean of all observations:

$\bar{x}=\frac{\bar{x}_{1}n_{1}+\bar{x}_{2}n_{2}+.....+\bar{x}_{n}n_{n}}{n_{1}+n_{2}+.......+n_{n}}=$ $\frac{\sum_{i=1}^{n}n_{i}\bar{x}_{i}}{\sum_{i=1}^{n}n_{i}}$

Therefore option (C) is correct.

Posted by

infoexpert23

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