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.