The score of all golfers for a particular course is normally distributed with a mean of 75 and a standard deviation of 10.5 . Suppose a random sample of 15 golfers is obtained. Describe the sampling distribution of the sample mean score of golfers (shape, mean, standard deviation). Shape: Mean: Standard Deviation (round to two decimals): What is the probability that a random sample of 15 golfers results in a mean score exceeds 70 ? Round to four decimals.

Step 1 ：The problem is asking for the sampling distribution of the sample mean score of golfers. The Central Limit Theorem states that if you have a population with mean \(\mu\) and standard deviation \(\sigma\) and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large (usually n > 30). If the population is normal, then the theorem holds true even for samples smaller than 30. Since in this case, the population is normally distributed, we can apply the Central Limit Theorem even though our sample size is less than 30.

Step 2 ：The mean of the sampling distribution is the same as the mean of the population. The standard deviation of the sampling distribution (often called the standard error) is the standard deviation of the population divided by the square root of the sample size n. In this case, the population mean (\(\mu\)) is 75, the population standard deviation (\(\sigma\)) is 10.5, and the sample size (n) is 15.

Step 3 ：The question also asks for the probability that a random sample of 15 golfers results in a mean score exceeds 70. This is a question of finding the probability that the sample mean is greater than a certain value. We can find this probability using the z-score formula: \(z = (X - \mu) / (\sigma / \sqrt{n})\) where X is the value we are interested in (70 in this case), \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and n is the sample size. The z-score tells us how many standard deviations the value is from the mean. We can then use a z-table to find the probability that a z-score is less than the calculated value, which gives us the probability that the sample mean is less than 70. To find the probability that the sample mean is greater than 70, we subtract this value from 1.

Step 4 ：Final Answer: The shape of the sampling distribution is Normal. The mean of the sampling distribution is 75. The standard deviation of the sampling distribution is approximately 2.71. The probability that a random sample of 15 golfers results in a mean score that exceeds 70 is approximately 0.9674. So, the answers are: Shape: Normal Mean: 75 Standard Deviation: \(\boxed{2.71}\) Probability that sample mean exceeds 70: \(\boxed{0.9674}\)