1. **State the problem:** We need to find the GPA cutoff for the highest 2.5% of students, given that GPAs are normally distributed with mean $\mu = 2.9$ and standard deviation $\sigma = 0.6$. 2. **Understand the problem:** The highest 2.5% corresponds to the 97.5th percentile of the normal distribution. 3. **Find the z-score for the 97.5th percentile:** From the standard normal distribution table, the z-score corresponding to 0.975 is approximately $z = 1.96$. 4. **Use the z-score formula:** $$x = \mu + z \sigma$$ 5. **Calculate the GPA cutoff:** $$x = 2.9 + 1.96 \times 0.6 = 2.9 + 1.176 = 4.076$$ 6. **Interpret the result:** The GPA cutoff for the highest 2.5% is approximately **4.08**, so students with GPAs of about 4.1 or higher are in the top 2.5%. **Final answer:** The GPA cutoff of the highest 2.5% is approximately $4.1$.