Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $104,000. This distribution follows the normal distribution with a standard deviation of $31,000.

Required:
If we select a random sample of 42 households, what is the standard error of the mean?
Note: Round your answer to the nearest whole number.

To calculate the **standard error of the mean** ($SE_{\bar{x}}$), we use the formula:

\[
SE_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]

Where:
- $\sigma = 31,000$ (the standard deviation),
- $n = 42$ (the sample size).

Let's compute this.

\[
SE_{\bar{x}} = \frac{31,000}{\sqrt{42}}
\]The standard error of the mean for a sample of 42 households is approximately **4,783**.

Would you like any further details or clarifications?

Here are five related questions:
1. How does sample size affect the standard error of the mean?
2. What does the standard error of the mean represent in the context of sampling?
3. How would the result change if the sample size increased to 100 households?
4. Why is the normal distribution assumption important in this problem?
5. How is the standard error different from the standard deviation?

**Tip:** A larger sample size generally reduces the standard error, providing a more precise estimate of the population mean.

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $104,000. This distribution follows the normal distribution with a standard deviation of $31,000. If we select a random sample of 42 households, what is the standard error of the mean? Note: Round your answer to the nearest whole number.

Learn how to calculate the standard error of the mean for life insurance data. Given a sample size of 42 households and a standard deviation of $31,000, this problem demonstrates the use of the standard error formula and the central limit theorem. Suitable for high school students in Grades 10-12.

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