Final answer:
The sampling distribution of the sample mean x-bar for the given scenario is normally distributed, with the same mean as the population mean μ, which is 182 mg/dl, and a standard error calculated as σ/√n, where σ is the population standard deviation and n is the sample size.
Step-by-step explanation:
The question asks about the sampling distribution of the sample mean (x-bar) for a normally distributed population of men's total cholesterol levels with a mean (μ) of 182 mg/dl and a standard deviation (s) of 37 mg. When drawing samples from a population where the population distribution is normal, the sampling distribution of the sample mean will also be normal. The mean of the sampling distribution will be the same as the population mean (μ = 182 mg/dl), and the standard deviation of the sampling distribution, known as the standard error (SE), will be the population standard deviation divided by the square root of the sample size (n). Therefore, if the sample size is not provided, we would use the formula SE = σ/√n to calculate the standard error.