1. **State the problem:** We are given a normal distribution of finishing times with mean $\mu = 69.8$ seconds and standard deviation $\sigma = 1.3$ seconds. We want to find the slowest time $x$ such that the RP men's team ranks in the top 10% of all times. This means their time must be less than or equal to the 10th percentile of the distribution. 2. **Understand the percentile:** The top 10% fastest times correspond to the lowest 10% of the finishing times because lower times are better. So we need to find the 10th percentile value $x$ where $P(X \leq x) = 0.10$. 3. **Find the z-score for the 10th percentile:** Using the standard normal distribution table or a calculator, the z-score $z$ for the 10th percentile is approximately $z = -1.28$. 4. **Use the z-score formula:** $$ z = \frac{x - \mu}{\sigma} $$ Substitute $z = -1.28$, $\mu = 69.8$, and $\sigma = 1.3$: $$ -1.28 = \frac{x - 69.8}{1.3} $$ 5. **Solve for $x$:** $$ x - 69.8 = -1.28 \times 1.3 = -1.664 $$ $$ x = 69.8 - 1.664 = 68.136 $$ 6. **Interpretation:** The slowest allowable time to be in the top 10% is approximately $68.14$ seconds. **Final answer:** $$\boxed{68.14 \text{ seconds}}$$