Question

If $\overline{x}$ and $\overline{y}$ are the means of two distributions such that $\overline{x}<\overline{y}$ and $\overline{z}$ is the mean of the combined distribution, then which one of the following statements is correct?

• A. $\overline{x}<\overline{y}<\overline{z}$
• B. $\overline{x}>\overline{y}>\overline{z}$
• C. $\overline{z}=\frac{\overline{x}+\overline{y}}{2}$
• D. $\overline{x}<\overline{z}<\overline{y}$

Answer 1

Answer: (D)

$\overline{x}<\overline{z}<\overline{y}$

Given $\overline{x}$ and $\overline{y}$ are two means of distributions and also $\overline{x}<\overline{y}$,$\overline{z}$ is the combined mean of the distribution.

Let $\overline{x}=\frac{S_1}{n_1}$,where $S_1$ is sum of the values in the distribution, and $n_1$ is the number of values in the distribution,

Similarly let,$\overline{y}=\frac{S_2}{n_2}$,

As $\overline{z}$ is the combined mean,

$⟹\overline{z}=\frac{S_1+S_2}{n_1+n_2}$

$⟹\overline{z}=\frac{n_1\overline{x}+n_2\overline{y}}{n_1+n_2}⟹\overline{z}=\frac{n_1\overline{x}}{n_1+n_2}+\frac{n_2\overline{y}}{n_1+n_2}$,

As $\frac{n_2\overline{y}}{n_1+n_2}$ is a positive number,

$⟹\overline{z}>\frac{n_1\overline{x}}{n_1+n_2}$,

As $n_1<n_1+n_2$,

$⟹\overline{z}>\overline{x}$,

As $\overline{y}>\overline{x}$,

$⟹\overline{z}<\overline{y}\frac{n_1}{n_1+n_2}+\overline{y}\frac{n_2}{n_1+n_2}⟹\overline{z}<\overline{y}\left(\frac{n_1+n_2}{n_1+n_2}\right)⟹\overline{z}<\overline{y}$

$\therefore$ $\overline{x}<\overline{z}<\overline{y}$