Final answer:
Using the normal distribution of PGA tour driving distances with a mean of 287.9 yards and a standard deviation of 8.4 yards, it is calculated that a driving distance at the 95th percentile is approximately 304.8 yards.
Step-by-step explanation:
The student is asking about determining a percentile value in a normally distributed data set, which is a mathematical concept. Since we know the mean (287.9 yards) and standard deviation (8.4 yards) for driving distances in the PGA tour, we can use z-scores to find the 95th percentile. First, we find the z-score that corresponds to the 95th percentile, which is typically around 1.645 for a one-tailed test. Next, we use the z-score formula:
Z = (X - μ) / σ,
Aiming to solve for X, where μ is the mean, σ is the standard deviation, and Z is the z-score. We then rearrange to:
X = Zσ + μ,
Substituting the values in, we get:
X = 1.645(8.4) + 287.9 = 304.786,
Therefore, the driving distance for the 95th percentile is approximately 304.8 yards.