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 5.50 years and a standard deviation of 2.20 years. Random samples of size 20 are drawn from the population and the mean of each sample is determined. Round the answers to the nearest hundredth.

A. 5.50 years, 0.49 years
B. 1.23 years, 2.20 years
C. 1.23 years, 0.49 years
D. 5.50 years, 0.11 years

Answer 1

The mean of the sampling distribution of employees' years of service is 5.50 years, which is the same as the population mean. The standard error of the mean is approximately 0.49 years, calculated by dividing the population standard deviation (2.20 years) by the square root of the sample size (20). option a is correct.

Step-by-step explanation:

When applying the Central Limit Theorem to find the mean and standard error of the mean of the indicated sampling distribution, we must first understand the given information. The population mean (μ) is 5.50 years and the population standard deviation (σ) is 2.20 years. Since random samples of size 20 are drawn, our sample size (n) is 20.

To find the sampling distribution mean, which is the same as the population mean according to the Central Limit Theorem, we use the given population mean directly:

• The mean (μ) of the sampling distribution is 5.50 years.

The standard error of the mean (SE), which measures the spread of all sample means around the population mean, is calculated by dividing the population standard deviation by the square root of the sample size:

SE = σ / √n

SE = 2.20 years / √20

SE = 2.20 years / 4.47

SE ≈ 0.49 years

Therefore, the correct answer to the question is Option A: The mean of the sampling distribution is 5.50 years and the standard error of the mean is 0.49 years.