25 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of 21.7 pounds. Assume that o=2.8 pounds. What Is the 95% confidence interval for the true mean weight, u, of all packages recelved by the parcel service? O20.6 to 22.8 pounds O 20.2 to 23.2 pounds O 20.7 to 22.7 pounds O 20.8 to 22.6 pounds

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### Determining the 95% Confidence Interval for Mean Weight of Packages

In this example, we have 25 packages that are randomly selected from packages received by a parcel service. The sample has a mean weight of 21.7 pounds. Assume that the standard deviation (σ) is 2.8 pounds.

**Question:**
What is the 95% confidence interval for the true mean weight, μ, of all packages received by the parcel service?

### Multiple Choice Options:
- \( \boxed{X} \) \( 20.6 \text{ to } 22.8 \text{ pounds} \)
- \( \boxed{} \) \( 20.2 \text{ to } 23.2 \text{ pounds} \)
- \( \boxed{} \) \( 20.7 \text{ to } 22.7 \text{ pounds} \)
- \( \boxed{} \) \( 20.8 \text{ to } 22.6 \text{ pounds} \)

To determine the correct confidence interval, we need to use the formula for the confidence interval for the mean, given by:

\[ \bar{x} \pm Z \left( \frac{\sigma}{\sqrt{n}} \right) \]

Where:
- \( \bar{x} \) is the sample mean
- \( σ \) is the population standard deviation
- \( n \) is the sample size
- \( Z \) is the Z-value from the standard normal distribution for the desired confidence level (for 95%, Z ≈ 1.96)
Transcribed Image Text:### Determining the 95% Confidence Interval for Mean Weight of Packages In this example, we have 25 packages that are randomly selected from packages received by a parcel service. The sample has a mean weight of 21.7 pounds. Assume that the standard deviation (σ) is 2.8 pounds. **Question:** What is the 95% confidence interval for the true mean weight, μ, of all packages received by the parcel service? ### Multiple Choice Options: - \( \boxed{X} \) \( 20.6 \text{ to } 22.8 \text{ pounds} \) - \( \boxed{} \) \( 20.2 \text{ to } 23.2 \text{ pounds} \) - \( \boxed{} \) \( 20.7 \text{ to } 22.7 \text{ pounds} \) - \( \boxed{} \) \( 20.8 \text{ to } 22.6 \text{ pounds} \) To determine the correct confidence interval, we need to use the formula for the confidence interval for the mean, given by: \[ \bar{x} \pm Z \left( \frac{\sigma}{\sqrt{n}} \right) \] Where: - \( \bar{x} \) is the sample mean - \( σ \) is the population standard deviation - \( n \) is the sample size - \( Z \) is the Z-value from the standard normal distribution for the desired confidence level (for 95%, Z ≈ 1.96)
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