Question

Question 17
If $\overline{x_1},\overline{x_2},\overline{x_3},\dots \dots \overline{x_n}$, are the means of n groups with $n_1,n_2,n_n$ number of observations, respectively. Then the mean $\overline{x}$ of all the groups:

A) ${\sum }_{i=1}^n{n}_i\overline{x_i}$
B) $\frac{{\sum }_{i=1}^n{n}_i\overline{x_i}}{n^2}$
C) $\frac{{\sum }_{i=1}^n{n}_i\overline{x_i}}{{\sum }_{i=1}^n{n}_i}$
D) $\frac{{\sum }^n{n}_i\overline{x_i}}{2n}$

Answer 1

The answer is C.
Given, $x_1,x_2\dots \dots x_n$ are the means of n groups having number of observations $n_1,n_2,\dots \dots n_n$ respectively.
Then, $n_1\overline{x_1}={\sum }_{i=1}^n{x}_i,n_2\overline{x_2}={\sum }_{i=1}^{n_2}{x}_j,n_3\overline{x_3}={\sum }_{k=1}^{n_3}{x}_k\dots \dots n_n\overline{x_n}={\sum }_{p=1}^{n_n}{x}_p$
Now, the mean $\overline{x}$ of all the groups taken together is given by
$\overline{x}=\frac{{\sum }_{i=1}^n{x}_i+{\sum }_{i=1}^{n_2}{x}_j+{\sum }_{k=1}^{n_3}{x}_k\dots \dots +{\sum }_{p=1}^{n_n}{x}_p}{n_1+n_2+n_n}$
$=\frac{n_1\overline{x_1}+n_2\overline{x_2}+n_3\overline{x_3}+\dots \dots +n_n\overline{x_n}}{n_1+n_2+\dots \dots +n_n}=\frac{{\sum }_{i=1}^n{n}_i\overline{x_i}}{{\sum }_{i=1}^n{n}_i}$
Hence, the mean of all the groups taken together is given by,
$\overline{x}=\frac{{\sum }_{i=1}^n{n}_i\overline{x_i}}{{\sum }_{i=1}^n{n}_i}$