1. **Problem Statement:** We have 49 fly balls with an average distance traveled, and we want to find the probability that the average distance is less than 222 feet. 2. **Given Information and Formula:** Assuming the distances are normally distributed with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample mean $\bar{X}$ for $n=49$ balls has: - Mean: $\mu_{\bar{X}} = \mu$ - Standard deviation (standard error): $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$ The probability we want is: $$P(\bar{X} < 222) = P\left(Z < \frac{222 - \mu}{\sigma/\sqrt{49}}\right)$$ where $Z$ is a standard normal variable. 3. **Assuming values:** Since the problem does not specify $\mu$ and $\sigma$, we infer from the graph scale (210 to 250) and typical baseball fly ball distances that: - $\mu = 230$ feet - $\sigma = 15$ feet 4. **Calculate the standard error:** $$\sigma_{\bar{X}} = \frac{15}{\sqrt{49}} = \frac{15}{7} = 2.1429$$ 5. **Calculate the Z-score:** $$Z = \frac{222 - 230}{2.1429} = \frac{-8}{2.1429} \approx -3.73$$ 6. **Find the probability:** Using standard normal tables or a calculator: $$P(Z < -3.73) \approx 0.0001$$ 7. **Interpretation:** The probability that the average distance of 49 balls is less than 222 feet is approximately 0.0001. --- **Part (c): Find the 70th percentile of the distribution of the average of 49 fly balls.** 1. The 70th percentile $P_{70}$ satisfies: $$P(\bar{X} < P_{70}) = 0.70$$ 2. Find the Z-score for 0.70: $$Z_{0.70} \approx 0.524$$ 3. Convert Z-score to $P_{70}$: $$P_{70} = \mu + Z_{0.70} \times \sigma_{\bar{X}} = 230 + 0.524 \times 2.1429 \approx 230 + 1.12 = 231.12$$ **Final answers:** - Probability that average is less than 222 feet: **0.0001** - 70th percentile of average distance: **231.12 feet** --- **Graph description:** The graph is a normal distribution curve with horizontal axis from 210 to 250. - The region to the left of 222 is shaded (top-left graph) representing the probability in part (b).