33.8k views
0 votes
If the sampled population has a mean of 48 and standard deviation of 16, then respectively the mean and the standard deviation for the sampling distribution of x¯ for n = 16 are?

1 Answer

4 votes

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.

User Jasonleakey
by
8.2k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories