A company’s marketing team takes 50 samples of 10 recent clients to create a sampling distribution of sample means for the average amount spent per month on company products. Can the Central Limit Theorem be used to determine that the sampling distribution is normal?
Table of contents
- 1. Intro to Stats and Collecting Data55m
- 2. Describing Data with Tables and Graphs1h 55m
- 3. Describing Data Numerically1h 45m
- 4. Probability2h 16m
- 5. Binomial Distribution & Discrete Random Variables2h 33m
- 6. Normal Distribution and Continuous Random Variables1h 38m
- 7. Sampling Distributions & Confidence Intervals: Mean1h 53m
- 8. Sampling Distributions & Confidence Intervals: Proportion1h 12m
- 9. Hypothesis Testing for One Sample2h 19m
- 10. Hypothesis Testing for Two Samples3h 26m
- 11. Correlation1h 6m
- 12. Regression1h 35m
- 13. Chi-Square Tests & Goodness of Fit1h 57m
- 14. ANOVA1h 0m
7. Sampling Distributions & Confidence Intervals: Mean
Sampling Distribution of the Sample Mean and Central Limit Theorem
Problem 5.4.24
Textbook Question
Interpreting the Central Limit Theorem In Exercises 19–26, find the mean and standard deviation of the indicated sampling distribution of sample means. Then sketch a graph of the sampling distribution.
Salaries The annual salaries for web software development managers are normally distributed, with a mean of about $136,000 and a standard deviation of about $11,500. Random samples of 40 are drawn from this population, and the mean of each sample is determined.

1
Identify the population mean (\$\mu\$) and population standard deviation (\$\sigma\$). Here, \$\mu = 136,000\$ and \$\sigma = 11,500\$.
Note the sample size \$n = 40\$ and recognize that the sampling distribution of the sample mean will have its own mean and standard deviation.
Calculate the mean of the sampling distribution of the sample mean, which is the same as the population mean: \$\mu_{\bar{x}} = \mu = 136,000\$.
Calculate the standard deviation of the sampling distribution of the sample mean (also called the standard error) using the formula: \$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{11,500}{\sqrt{40}}\$.
Since the population distribution is normal, the sampling distribution of the sample mean will also be normal. Sketch a normal curve centered at \$136,000\$ with spread determined by \$\sigma_{\bar{x}}\$.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Central Limit Theorem (CLT)
The Central Limit Theorem states that the sampling distribution of the sample mean approaches a normal distribution as the sample size becomes large, regardless of the population's distribution. For sample sizes typically greater than 30, the sample means will be approximately normally distributed, enabling inference about the population mean.
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Sampling Distribution of the Sample Mean
The sampling distribution of the sample mean is the probability distribution of all possible sample means from samples of a fixed size. It has a mean equal to the population mean and a standard deviation (standard error) equal to the population standard deviation divided by the square root of the sample size.
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Standard Error of the Mean
The standard error measures the variability of the sample mean from the population mean. It is calculated as the population standard deviation divided by the square root of the sample size, reflecting how much sample means fluctuate around the true mean in repeated sampling.
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