1. **Problem Statement:** We have fly ball distances normally distributed with mean $\mu=234$ feet and standard deviation $\sigma=56$ feet. We sample $n=49$ fly balls and want to find the distribution of the sample mean $\bar{X}$ and the probability that $\bar{X}<222$ feet. 2. **Distribution of Sample Mean:** The sample mean $\bar{X}$ of $n$ independent normal variables is also normally distributed with mean $\mu_{\bar{X}}=\mu=234$ and standard deviation $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{56}{\sqrt{49}}=8$. So, $$\bar{X} \sim N(234, 8)$$ 3. **Calculating the Probability $P(\bar{X}<222)$:** We standardize the value 222 to a $z$-score: $$z=\frac{222-234}{8} = \frac{-12}{8} = -1.5$$ 4. **Using the Standard Normal Distribution Table:** The probability that $Z < -1.5$ is approximately 0.0668. Therefore, $$P(\bar{X} < 222) = 0.0668$$ 5. **Interpretation:** There is about a 6.68% chance that the average distance of 49 fly balls is less than 222 feet. 6. **Graph Sketch Explanation:** The graph shows a normal curve centered at 234 with the area to the left of 222 shaded, representing the probability $P(\bar{X}<222)$. Final answers: - Distribution: $$\bar{X} \sim N(234, 8)$$ - Probability: $$P(\bar{X} < 222) = 0.0668$$