Final answer:
The 60th percentile of the distribution of fly balls, with a mean of 250 feet and a standard deviation of 50 feet, is found by identifying the z-score for the 60th percentile (approximately 0.25) and applying it to the z-score formula. The 60th percentile is 262.5 feet.
Step-by-step explanation:
To find the 60th percentile of the distribution of fly balls in baseball, we use the properties of the normal distribution. Given that the mean distance (μ) is 250 feet and the standard deviation (σ) is 50 feet, we look for a z-score that corresponds to the 60th percentile.
First, we consult the standard normal distribution table (or use a calculator or software) to find the z-score that corresponds to the 60th percentile, which is approximately 0.25. Then, we use the z-score formula to convert this to the distance X for the fly balls:
Z = (X - μ) / σ
Solving for X, we have:
X = Z×σ + μ
X = 0.25× 50 + 250
X = 12.5 + 250
X = 262.5 feet
Therefore, the 60th percentile of the distribution of fly balls is 262.5 feet