1. **Problem statement:** We need to find the new minimum height for men such that only the tallest 4% of men are eligible. 2. **Given data:** - Men's heights are normally distributed with mean $\mu = 69.0$ inches and standard deviation $\sigma = 2.8$ inches. - We want the cutoff height $h$ such that the top 4% of men are taller than $h$. 3. **Formula and concept:** We use the standard normal distribution $Z = \frac{X - \mu}{\sigma}$. We want to find $h$ such that $P(X > h) = 0.04$. This means $P(X \leq h) = 0.96$. 4. **Find the z-score for 0.96 cumulative probability:** From standard normal tables or using inverse normal function, $z_{0.96} \approx 1.75$. 5. **Calculate the cutoff height:** $$h = \mu + z \times \sigma = 69.0 + 1.75 \times 2.8 = 69.0 + 4.9 = 73.9$$ 6. **Interpretation:** The new minimum height for men to be in the tallest 4% is approximately 73.9 inches. **Final answer:** $$\boxed{73.9 \text{ inches}}$$