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 B $\overline{x}>{\overline{x}}_2$

Let $n_1$ and $n_2$ be the number of observations in two groups having means $\overline{x_1}$ and $\overline{x_2}$
Respectively. Then,$\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}$
$Now,\overline{x}-{\overline{x}}_1=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}-{\overline{x}}_1$
$=\frac{n_2({\overline{x}}_2-{\overline{x}}_1)}{n_1+n_2}=0,[\because {\overline{x}}_2>{\overline{x}}_1]$
$\Rightarrow \overline{x}>{\overline{x}}_1.....\left(i\right)$
$And\overline{x}>{\overline{x}}_2=\frac{n({\overline{x}}_1-{\overline{x}}_2)}{n_1+n_2}<0,[\because {\overline{x}}_2>{\overline{x}}_1]$
$\Rightarrow \overline{x}>{\overline{x}}_2....\left(ii\right)$
$From\left(i\right)and\left(ii\right){\overline{x}}_1<\overline{x}<{\overline{x}}_2.$