1. **Problem statement:** Find the 70th percentile of the distribution of the average of 49 fly balls. 2. **Given:** The sample size $n=49$. 3. **Recall:** The distribution of the sample mean $\bar{X}$ for a sample size $n$ from a population with mean $\mu$ and standard deviation $\sigma$ has mean $\mu_{\bar{X}} = \mu$ and standard deviation (standard error) $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$. 4. **Find the z-score for the 70th percentile:** Using standard normal distribution tables or a calculator, the z-score $z_{0.70}$ corresponding to the 70th percentile is approximately $0.524$. 5. **Calculate the 70th percentile of the sample mean distribution:** $$ P_{70} = \mu + z_{0.70} \times \frac{\sigma}{\sqrt{n}} $$ 6. **Substitute values:** Since the previous problem's mean $\mu$ and standard deviation $\sigma$ are not given here, assume you have them from the previous problem. For example, if $\mu=\mu_0$ and $\sigma=\sigma_0$, then $$ P_{70} = \mu_0 + 0.524 \times \frac{\sigma_0}{7} $$ 7. **Round the result to 2 decimal places.** **Note:** Please replace $\mu_0$ and $\sigma_0$ with the actual values from the previous problem to compute the numerical answer.