On a test that has a normal distribution of scores of 59 falls two standard deviations below the mean, and score of 74 is one standard deviation above the mean. If x is an integer score that lies between 2.5 and 3.0 standard deviations above the means. What is a possible value of x?

Answer

To solve the problem, we need to find the integer score \( x \) that lies between 2.5 and 3.0 standard deviations above the mean, given the scores of 59 and 74 in relation to the mean and standard deviation.

### Step-by-Step Solution:

1. **Define the Mean and Standard Deviation**:
Let the mean be represented by \( \mu \) and the standard deviation by \( \sigma \).

2. **Set Up the Equations**:
- From the information that 59 falls two standard deviations below the mean, we can write:
\[ \mu - 2\sigma = 59 \quad \text{(1)} \]
- From the information that 74 is one standard deviation above the mean, we can write:
\[ \mu + \sigma = 74 \quad \text{(2)} \]

3. **Solve the Equations**:
- We can subtract equation (1) from equation (2):
\[ (\mu + \sigma) - (\mu - 2\sigma) = 74 - 59 \]
Simplifying this gives:
\[ 3\sigma = 15 \]
Therefore,
\[ \sigma = \frac{15}{3} = 5 \]

4. **Find the Mean**:
- Now, substitute \( \sigma = 5 \) back into equation (2) to find \( \mu \):
\[ \mu + 5 = 74 \]
Thus,
\[ \mu = 74 - 5 = 69 \]

5. **Determine the Range for \( x \)**:
- We need to find the range for \( x \) which lies between 2.5 and 3.0 standard deviations above the mean:
\[ \mu + 2.5\sigma < x < \mu + 3.0\sigma \]
- Substitute the values of \( \mu \) and \( \sigma \):
\[ 69 + 2.5 \times 5 < x < 69 + 3.0 \times 5 \]
This simplifies to:
\[ 69 + 12.5 < x < 69 + 15 \]
Therefore:
\[ 81.5 < x < 84 \]

6. **Identify Possible Integer Values**:
- The integers that lie between 81.5 and 84 are 82 and 83.

### Conclusion:
Thus, the possible integer values of \( x \) are **82 and 83**.