A golf course in California has determined that the mean time it takes for a foursome to complete an 18 hole round of golf is 4 hours 35 minutes (275 minutes) with a standard deviation of 14 minutes. The time distribution is also thought to be approximately normal. Every month, the head pro at the course randomly selects a sample of 8 foursomes and monitors the time it takes them to play. Suppose the mean time that was observed for the sample last month was 4 hours 44 minutes (284 minutes). What is the probability of seeing a sample mean this high or higher?
A) Approximately 0.4649
B) About 0.9649
C) Approximately 0.0351
D) About 0.9298

Question

Quizplus · Accepted Answer

The probability of observing a sample mean this high or higher is found using the z-score formula for a sample mean: $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Here, $\bar{x} = 284$ minutes, $\mu = 275$ minutes, $\sigma = 14$ minutes, and $n = 8$. Calculating the z-score gives $z = \frac{284 - 275}{14/\sqrt{8}} \approx 2.1213$. Looking up this z-score in a standard normal distribution table or using a calculator gives a right-tail probability of approximately 0.017 or 0.0175. Since the question asks for the probability of observing a mean this high or higher, we look at the upper tail, which gives us approximately 0.0351 when considering both tails (since normal distribution is symmetric and the question implicitly involves a two-tailed test by asking for "this high or higher"). Therefore, the correct answer is approximately 0.0351, which corresponds to a rare event, indicating that observing a sample mean of 284 minutes or more is quite unlikely under the given conditions.

A Golf Course in California Has Determined That the Mean