You take a random sample of 36 cantaloupe weights. The sample has a mean of 2.5 lbs and a standard deviation of 0.4 pounds. The population appears to be normally distributed. If you construct a 95% confidence interval of the population mean, what is the margin of error?

To find the **margin of error** for the 95% confidence interval, we can use the formula for the margin of error (ME) when dealing with a normally distributed population:

\[
ME = z \cdot \frac{\sigma}{\sqrt{n}}
\]

Where:
- \( z \) is the critical value corresponding to the confidence level (for 95%, \( z = 1.96 \)),
- \( \sigma \) is the standard deviation of the sample (0.4 lbs),
- \( n \) is the sample size (36).

Now let's plug in the values:

\[
ME = 1.96 \cdot \frac{0.4}{\sqrt{36}} = 1.96 \cdot \frac{0.4}{6} = 1.96 \cdot 0.0667 \approx 0.1307
\]

So, the **margin of error** is approximately **0.13 pounds**.

Would you like more details or have any questions about this? Here are five related questions for further exploration:

1. How would the margin of error change if the sample size increased?
2. What is the full 95% confidence interval for the population mean in this case?
3. What is the critical value \( z \) for a 99% confidence interval?
4. How does the margin of error relate to the confidence level chosen?
5. How does a larger standard deviation affect the margin of error?

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

Learn how to calculate the margin of error for a 95% confidence interval using a sample of 36 cantaloupe weights with a mean of 2.5 lbs and a standard deviation of 0.4 lbs. Understand the statistical concepts involved, including confidence intervals and the normal distribution. Ideal for Grades 10-12.

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