Suppose the average driving distance for last year's Masters Golf Tournament was 282.5 yards with a standard deviation of 12.2 yards. A random sample of 30 drives was selected from a total of 4,144 drives that were hit during this tournament. What is the probability that the sample average was 279 yards or less?
A)0.3540
B)0.1611
C)0.0582
D)0.2676

Question

Quizplus · Accepted Answer

To find the probability that the sample average is 279 yards or less, we use the standard normal distribution. First, calculate the standard error (SE) using the formula SE = σ/√n, where σ is the standard deviation and n is the sample size. SE = 12.2/√30 ≈ 2.23. Then, calculate the z-score using the formula z = (X̄ - μ) / SE, where X̄ is the sample mean, and μ is the population mean. z = (279 - 282.5) / 2.23 ≈ -1.57. Using the standard normal distribution table, the probability corresponding to z = -1.57 is approximately 0.0582.

Suppose the Average Driving Distance for Last Year's Masters Golf