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 combined distribution, then

(Given : The two distributions are of equal length)

• 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 $\overline{X_1}<\overline{X}<\overline{X_2}$

If $\overline{X_1}and\overline{X_2}$ are the means of two distributions such that $\overline{X_1}<\overline{X_2}$
$\overline{X}$ is the mean of combined distribution
$\overline{X}=\frac{\overline{X_1}+\overline{X_2}}{2}$
$\overline{X_1}<\overline{X_2}$ $\overline{X_1}<\overline{X_2}$
$2\overline{X_1}<\overline{X_1}+\overline{X_2}$ $\overline{X_1}+\overline{X_2}<2\overline{X_2}$
$\overline{X_1}<\frac{\overline{X_1}+\overline{X_2}}{2}$ $\frac{\overline{X_1}+\overline{X_2}}{2}<\overline{X_2}$
$\overline{X_1}<\overline{X}$ $\overline{X}<\overline{X_2}$
Adding both equation, we get
$\overline{X_1}<\overline{X}<\overline{X_2}$