Final answer:
The sampling distribution of the sample mean will be approximately normal according to the Central Limit Theorem, with the distribution's mean equal to the population mean and its standard deviation equal to the population's divided by the square root of the sample size. You'll use the standard normal distribution for calculating probabilities related to the sample means.
Step-by-step explanation:
The central concept here is the Central Limit Theorem (CLT), which states that the distribution of sample means will tend to be normal, or bell-shaped, as the sample size increases, regardless of the population's initial distribution. When dealing with a sufficiently large sample size, the mean of the sampling distribution of the sample mean (μx) will be equal to the population mean (μ), while the standard deviation of the sampling distribution (σx) is equal to the population standard deviation (σ) divided by the square root of the sample size (n).
Applying this to a specific example, for a population with a mean (μ) of 50 and a standard deviation (σ) of 4, when you draw samples of size 40, the sampling distribution will have a mean (μx) of 50 and a standard deviation (σx) of 4/√40. According to the CLT, the shape of this distribution will be approximately normal for large n. Therefore, most sample means will cluster around μ, with the spread determined by σx.
For calculating probabilities such as P(xbar > a certain value), you would use the standard normal distribution because the sampling distribution is normal. For example, P(xbar > 73), find the z-score by subtracting the mean of the sampling distribution from 73 and dividing by the standard deviation of the sampling distribution, and then consult a z-table or use statistical software to find the corresponding probability.