Let `barx` be the mean of `x_(1), x_(2), ………, x_(n)` and `bary` be the mean of `y_(1), y_(2),……….,y_(n)`.
If `barz` is the mean of `x_(1), x_(2), ……………..x_(n), y_(1), y_(2), …………,y_(n)`, then `barz` is equal to

Options

A. `barx+bary`
B. `(barx+bary)/(2)`
C. `(barx+bary)/n`
D. `(barx+bary)/(2n)`

Answer

To solve the problem, we need to find the mean \( \bar{z} \) of the combined observations \( x_1, x_2, \ldots, x_n, y_1, y_2, \ldots, y_n \) based on the given means \( \bar{x} \) and \( \bar{y} \).

### Step-by-Step Solution:

1. **Understand the Means**:
- The mean \( \bar{x} \) of the observations \( x_1, x_2, \ldots, x_n \) is defined as:
\[ \bar{x} = \frac{x_1 + x_2 + \ldots + x_n}{n} \]
- Therefore, the sum of these observations can be expressed as:
\[ x_1 + x_2 + \ldots + x_n = n \bar{x} \quad \text{(Equation 1)} \]

2. **Calculate the Mean of \( y \)**:
- Similarly, the mean \( \bar{y} \) of the observations \( y_1, y_2, \ldots, y_n \) is defined as:
\[ \bar{y} = \frac{y_1 + y_2 + \ldots + y_n}{n} \]
- Thus, the sum of these observations can be expressed as:
\[ y_1 + y_2 + \ldots + y_n = n \bar{y} \quad \text{(Equation 2)} \]

3. **Combine the Observations**:
- The combined observations are \( x_1, x_2, \ldots, x_n, y_1, y_2, \ldots, y_n \). The total number of observations is \( n + n = 2n \).

4. **Calculate the Mean \( \bar{z} \)**:
- The mean \( \bar{z} \) of all these observations is given by:
\[ \bar{z} = \frac{x_1 + x_2 + \ldots + x_n + y_1 + y_2 + \ldots + y_n}{2n} \]
- Substituting the sums from Equations 1 and 2 into this expression:
\[ \bar{z} = \frac{n \bar{x} + n \bar{y}}{2n} \]

5. **Simplify the Expression**:
- We can factor out \( n \) from the numerator:
\[ \bar{z} = \frac{n(\bar{x} + \bar{y})}{2n} \]
- The \( n \) in the numerator and denominator cancels out:
\[ \bar{z} = \frac{\bar{x} + \bar{y}}{2} \]

### Final Result:
Thus, the mean \( \bar{z} \) is given by:
\[ \bar{z} = \frac{\bar{x} + \bar{y}}{2} \]