If `x _(1) and x _(2) ` are the means of two distributions such that `x _(1) lt x _(2) and bar x` is the mean of the combined distribution, then

Options

A. `bar x lt bar x _(1)`
B. `bar x gt bar x _(2)`
C. `bar x = ( bar x _(1) + bar x _(2))/(2)`
D. `bar x _(1) lt bar x lt bar x _(2)`

Answer

To solve the given problem, we need to analyze the relationship between the means of two distributions, \( x_1 \) and \( x_2 \), and the mean of the combined distribution, denoted as \( \bar{x} \).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We are given that \( x_1 < x_2 \) and that \( x_1 \) is the mean of the combined distribution. This means that the combined mean \( \bar{x} \) is equal to \( x_1 \).

2. **Expressing the Combined Mean**:
The combined mean \( \bar{x} \) of two distributions with means \( x_1 \) and \( x_2 \) can be expressed as:
\[ \bar{x} = \frac{x_1 + x_2}{2} \]
Since we know \( \bar{x} = x_1 \), we can set up the equation:
\[ x_1 = \frac{x_1 + x_2}{2} \]

3. **Multiplying Both Sides by 2**:
To eliminate the fraction, we multiply both sides of the equation by 2:
\[ 2x_1 = x_1 + x_2 \]

4. **Rearranging the Equation**:
Now, we can rearrange the equation to isolate \( x_2 \):
\[ 2x_1 - x_1 = x_2 \]
Simplifying this gives:
\[ x_2 = x_1 \]

5. **Analyzing the Inequality**:
However, we were given that \( x_1 < x_2 \). This means that our earlier assumption must be checked. Since \( x_1 \) is the mean of the combined distribution, it must be less than the mean of the second distribution \( x_2 \).

6. **Final Conclusion**:
Thus, we can conclude that:
\[ x_1 < x_2 \]
This confirms that \( x_1 \) is less than \( x_2 \), and since \( x_1 \) is the mean of the combined distribution, it is consistent with the information provided.

### Final Result:
The relationship we derived is:
\[ x_1 < x_2 \]