Question

If ${\overline{x}}_1$ and ${\overline{x}}_2$ are the means of two distributions such that ${\overline{x}}_1<{\overline{x}}_2$ and $\overline{x}$ is the mean of the combined distribution, then

• A. $\overline{x}<{\overline{x}}_1$
• B. $\overline{x}>{\overline{x}}_2$
• C. $\overline{x}=\frac{{\overline{x}}_1+{\overline{x}}_2}{2}$
• D. ${\overline{x}}_1<\overline{x}<{\overline{x}}_2$

Answer 1

The correct option is D ${\overline{x}}_1<\overline{x}<{\overline{x}}_2$

Let $n_1,n_2$ be the number of observations in distributions having means ${\overline{x}}_1,{\overline{x}}_2$ respectively.
Now, combined mean
$\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}$
$\because {\overline{x}}_1<{\overline{x}}_2$
$\Rightarrow n_1{\overline{x}}_1<n_1{\overline{x}}_2$
So,$\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}<\frac{n_1{\overline{x}}_2+n_2{\overline{x}}_2}{n_1+n_2}$
$\Rightarrow \overline{x}<{\overline{x}}_2\left(\frac{n_1+n_2}{n_1+n_2}\right)$
$\Rightarrow \overline{x}<{\overline{x}}_2...........\left(i\right)$
Also, ${\overline{x}}_2>{\overline{x}}_1$
$\Rightarrow n_2{\overline{x}}_2>n_2{\overline{x}}_1$
So, $\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}>\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_1}{n_1+n_2}$
$\Rightarrow \overline{x}>{\overline{x}}_1\left(\frac{n_1+n_2}{n_1+n_2}\right)$
$\Rightarrow \overline{x}>{\overline{x}}_1...........\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
${\overline{x}}_1<\overline{x}<{\overline{x}}_2$