Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 260 feet and a standard deviation of 38 feet. Let $X$ be the distance in feet for a fly ball. a. What is the distribution of $X ? X \sim N(260$ 38 b. Find the probability that a randomly hit fly ball travels less than 293 feet. Round to 4 decimal places. c. Find the 70th percentile for the distribution of distance of fly balls. Round to 2 decimal places. feet

Step 1 ：Let's denote the distance of fly balls hit to the outfield as \(X\). The distribution of \(X\) is normal with a mean of 260 feet and a standard deviation of 38 feet. So, we can write it as \(X \sim N(260, 38)\).

Step 2 ：We are asked to find the probability that a randomly hit fly ball travels less than 293 feet. This is equivalent to finding the cumulative probability \(P(X < 293)\) under the normal distribution.

Step 3 ：To find this probability, we first need to standardize the value 293 to a z-score. The z-score gives us the number of standard deviations away from the mean. We can calculate the z-score using the formula \(z = \frac{X - \mu}{\sigma}\), where \(\mu\) is the mean and \(\sigma\) is the standard deviation.

Step 4 ：Substituting the given values into the formula, we get \(z = \frac{293 - 260}{38} \approx 0.868421052631579\).

Step 5 ：Finally, we use the z-score to find the cumulative probability. Looking up the z-score in a standard normal distribution table or using a statistical software, we find that \(P(X < 293) \approx 0.8074180630347814\).

Step 6 ：So, the probability that a randomly hit fly ball travels less than 293 feet is approximately \(\boxed{0.8074}\).