Question

the weights of people in a certain population are normally distributed with a mean of 157 lb and a standard deviation of 24 lb. find the mean and standard error of the mean for this sampling distribution when using random samples of size 5. round the answers to the nearest hundredth.

a. 157.00, 5.00
b. 157.00, 4.80
c. 157.00, 24.00
d. 157.00, 10.73

Explanation:

Step1: Find the mean of the sampling distribution

The mean of the sampling distribution of the sample mean \(\mu_{\bar{x}}\) is equal to the population mean \(\mu\). Given \(\mu = 157\) lb, so \(\mu_{\bar{x}}=\mu = 157\) lb.

Step2: Find the standard error of the mean

The formula for the standard error of the mean \(\sigma_{\bar{x}}\) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size. Given \(\sigma = 24\) lb and \(n = 5\). Then \(\sigma_{\bar{x}}=\frac{24}{\sqrt{5}}\approx\frac{24}{2.24}\approx10.73\) (rounded to the nearest hundredth).

Answer:

D. \(157.00, 10.73\)