Final answer:
For a population with a mean of 48 and a standard deviation of 16, and a sample size of 16, the mean and standard deviation of the sampling distribution of the sample mean are 48 and 4, respectively.
Step-by-step explanation:
The question asks about the properties of the sampling distribution of the sample mean (notated as \(\bar{x}\)) given a population mean of 48, a population standard deviation of 16, and a sample size (n) of 16. According to the central limit theorem, the sampling distribution of the sample mean will have a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.
The mean of the sampling distribution (\(\mu_{\bar{x}}\)) is the same as the population mean, hence \(\mu_{\bar{x}} = \mu = 48\).
The standard deviation of the sampling distribution (\(\sigma_{\bar{x}}\)), also known as the standard error, is calculated as the population standard deviation divided by the square root of the sample size. Using the formula \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\), the standard deviation of the sampling distribution for this example is \(\sigma_{\bar{x}} = \frac{16}{\sqrt{16}} = \frac{16}{4} = 4\).
Therefore, for a sample size of 16, the mean and standard deviation of the sampling distribution of the sample mean (\(\bar{x}\)) are 48 and 4, respectively.