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. $\overline{x}_{1}<\overline{x}<\overline{x}_{2}$

Hint:

If $x1$ and $\overlinex2$ are the means of two distributions such that $\overlinex1<\overlinex2$ and $\overlinex$ is the mean of the combined distribution, then

Answer 1

The Correct Option is D

Step 1: Concept
The combined mean $\overline{x}$ of two groups with sizes $n_{1}, n_{2}$ and means $\overline{x}_{1}, \overline{x}_{2}$ is given by $\overline{x} = \frac{n_{1}\overline{x}_{1} + n_{2}\overline{x}_{2}}{n_{1} + n_{2}}$.
Step 2: Analysis
Subtracting $\overline{x}_{1}$ from the combined mean: $\overline{x} - \overline{x}_{1} = \frac{n_{2}(\overline{x}_{2} - \overline{x}_{1})}{n_{1} + n_{2}}$. Since $\overline{x}_{2}>\overline{x}_{1}$, this value is positive, so $\overline{x}>\overline{x}_{1}$.
Step 3: Analysis
Subtracting $\overline{x}_{2}$ from the combined mean: $\overline{x} - \overline{x}_{2} = \frac{n_{1}(\overline{x}_{1} - \overline{x}_{2})}{n_{1} + n_{2}}$. Since $\overline{x}_{1}<\overline{x}_{2}$, this value is negative, so $\overline{x}<\overline{x}_{2}$.
Step 4: Conclusion
Therefore, the combined mean always lies between the two individual means: $\overline{x}_{1}<\overline{x}<\overline{x}_{2}$.
Final Answer: (d)