A sample of 121 checked bags at an airport had an average weight of 37 pounds with a standard deviation of 7 pounds. (a) At 95% confidence, compute the margin of error (in pounds). (Round your answer to four decimal places.) lbs Explain what it shows. Approximately 95% of all samples of size 121 will produce a sample mean and margin of error such that the distance between the sample mean and the population mean is at most the margin of error. Approximately 95% of all samples of size 121 will produce a sample mean and margin of error such that the distance between the sample mean and the population mean is equal to the margin of error. We can say with 0.95 probability that the distance between the sample mean of 37 pounds and the population mean is at least the margin of error calculated above. We can say with 0.95 probability that the distance between the sample mean of 37 pounds and the population mean is at most the margin of error calculated above. Approximately 95% of all samples of size 121 will produce a sample mean and margin of error such that the distance between the sample mean and the population mean is at least the margin of error. (b) Determine a 95% confidence interval for the population mean weight of checked bags at the airport (in pounds). (Round your answers to two decimal places.)

Let's walk through both parts of the problem.

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### **(a) Margin of Error Calculation (95% Confidence Level)**  
The formula for the **margin of error (ME)** for a sample mean is:

\[
ME = Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}
\]

where:  
- \( Z_{\alpha/2} \) is the critical value for a 95% confidence level (1.960 for 95% confidence),
- \( \sigma \) is the standard deviation of the sample (7 pounds),
- \( n \) is the sample size (121 bags).  

Now, substitute the values:

\[
ME = 1.960 \times \frac{7}{\sqrt{121}}
\]

\[
ME = 1.960 \times \frac{7}{11} = 1.960 \times 0.6364 \approx 1.2467 \, \text{pounds}
\]

So, the **margin of error** is:

\[
\boxed{1.2467 \, \text{pounds}}
\]

---

#### **Explanation of the Margin of Error:**

Out of the provided options, the **correct interpretation** of the margin of error at 95% confidence is:

**"We can say with 0.95 probability that the distance between the sample mean of 37 pounds and the population mean is at most the margin of error calculated above."**

This means that with 95% confidence, the true population mean is within the range determined by adding and subtracting the margin of error from the sample mean.

---

### **(b) 95% Confidence Interval for the Population Mean**  

The **formula for the confidence interval** is:

\[
\text{CI} = \bar{X} \pm ME
\]

where:  
- \( \bar{X} = 37 \) pounds (sample mean),
- \( ME = 1.2467 \) pounds (calculated above).  

Now, calculate the lower and upper bounds of the interval:

\[
\text{Lower bound} = 37 - 1.2467 = 35.75
\]
\[
\text{Upper bound} = 37 + 1.2467 = 38.25
\]

So, the **95% confidence interval** for the population mean is:

\[
\boxed{(35.75, 38.25)} \, \text{pounds}
\]

---

### Summary

- **(a)** Margin of error: **1.2467 pounds**
- **(b)** 95% confidence interval: **(35.75, 38.25) pounds**

Would you like further details on any part of the calculation?

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Here are 5 related questions for deeper exploration:  
1. How would the margin of error change if the sample size increased to 400?  
2. What is the Z-critical value for a 99% confidence level, and how does it affect the margin of error?  
3. How do we interpret a confidence interval that includes the value 0?  
4. If the standard deviation was 10 pounds instead of 7, how would the margin of error change?  
5. What assumptions must hold for the confidence interval to be valid?

**Tip:** Increasing the sample size decreases the margin of error, making the estimate more precise.

A sample of 121 checked bags at an airport had an average weight of 37 pounds with a standard deviation of 7 pounds. (a) At 95% confidence, compute the margin of error (in pounds). (Round your answer to four decimal places.) Explain what it shows. (b) Determine a 95% confidence interval for the population mean weight of checked bags at the airport (in pounds). (Round your answers to two decimal places.)

Learn how to compute the margin of error and 95% confidence interval for the average weight of 121 checked bags at an airport, using a sample mean of 37 pounds and standard deviation of 7 pounds. This step-by-step guide explains statistical concepts such as margin of error, confidence intervals, and the normal distribution, and is suitable for advanced high school and undergraduate students.

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