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Question:

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

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

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

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

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

 

Solution:

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

Sum of all terms $=n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}+\ldots n_{n} \bar{x}_{n}$

Number of terms $=\left(n_{1}+n_{2}+\ldots n_{n}\right)$

$\therefore$ Mean $=\frac{\sum_{i=1}^{n} n_{i} \bar{x}_{i}}{\sum_{i=1}^{n} n_{i}}$

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