1. The problem asks for the probability that the average distance traveled by 49 balls is less than 222 feet. 2. We use the Central Limit Theorem which states that the sampling distribution of the sample mean will be approximately normal if the sample size is large enough. 3. The formula for the z-score of the sample mean is $$z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}$$ where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. 4. Assuming from the last problem the population mean $\mu = 224$ feet and population standard deviation $\sigma = 15$ feet. 5. Calculate the standard error: $$SE = \frac{15}{\sqrt{49}} = \frac{15}{7} = 2.1429$$ 6. Calculate the z-score for $\bar{x} = 222$: $$z = \frac{222 - 224}{2.1429} = \frac{-2}{2.1429} \approx -0.9333$$ 7. Using the standard normal distribution table or a calculator, find the probability corresponding to $z = -0.9333$. 8. The cumulative probability for $z = -0.9333$ is approximately 0.1759. 9. Therefore, the probability that the average distance traveled by 49 balls is less than 222 feet is approximately 0.1759. Final answer: **0.1759**