Suppose that grade point averages of undergraduate students at one university have a bell-shaped distribution with a mean of 2.58 and a standard deviation of 0.45. Please do not round your answer. describe where the highest and lowest 32% of grade average lie

To find the highest and lowest 32% of grade averages in a normal distribution:

Given:
\mu = 2.58
\sigma = 0.45

1. **Calculate the z-score for the 68th percentile:**

z = 0.47

2. **Convert this z-score to an actual grade point average for the highest 32%:**

x = \mu + (z \times \sigma)

x = 2.58 + (0.47 \times 0.45)

x = 2.58 + 0.2115

x = 2.7915

Thus, the highest 32% of grade point averages are above:

x > 2.7915

3. **Calculate the z-score for the 32nd percentile:**

z = -0.47

4. **Convert this z-score to an actual grade point average for the lowest 32%:**

x = \mu + (z \times \sigma)

x = 2.58 + (-0.47 \times 0.45)

x = 2.58 - 0.2115

x = 2.3685

Thus, the lowest 32% of grade point averages are below:

x < 2.3685

**Summary:**

- The highest 32% of grade point averages lie above:

x > 2.7915

- The lowest 32% of grade point averages lie below:

x < 2.3685

**Answer:**
- \text{Highest 32\%: } x > 2.7915
- \text{Lowest 32\%: } x < 2.3685