Question

use the central limit theorem to find the mean and standard error of the mean of the indicated sampling distribution. the amounts of time employees of a telecommunications company have worked for the company are normally distributed with a mean of 5.80 years and a standard deviation of 2.10 years. random samples of size 17 are drawn from the population and the mean of each sample is determined. round the answers to the nearest hundredth. oa 1.41 years, 2.10 years ob. 5.80 years, 0.51 years oc. 5.80 years, 0.12 years od. 1.41 years, 0.51 years

Explanation:

Step1: Find the mean of the sampling distribution

According to the Central Limit Theorem, if the population has mean \(\mu\), then the mean of the sampling distribution of the sample mean \(\bar{X}\), denoted as \(\mu_{\bar{X}}\), is equal to the population mean. Given \(\mu = 5.80\) years, so \(\mu_{\bar{X}}=\mu = 5.80\) years.

Step2: Find the standard error of the sampling distribution

The formula for the standard error (standard deviation of the sampling distribution of the sample mean) is \(\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size. Given \(\sigma = 2.10\) years and \(n = 17\). Then \(\sigma_{\bar{X}}=\frac{2.10}{\sqrt{17}}\approx\frac{2.10}{4.123}\approx0.51\) years.

Answer:

B. \(5.80\) years, \(0.51\) years