The standard deviation of test scores on a certain achievement is 10.6. A random sample of 90 scores on this test had a mean of 75.9. Based on this sample, find a 99% confidence interval for the true mean of all scores. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place.

To find a 99% confidence interval for the true mean of all scores based on the sample data provided, we'll use the formula for the confidence interval of the mean when the population standard deviation is known. The formula is:

\[
\text{CI} = \bar{x} \pm Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}
\]

Where:
- \(\bar{x}\) is the sample mean.
- \(Z_{\alpha/2}\) is the critical value from the standard normal distribution for a 99% confidence level.
- \(\sigma\) is the population standard deviation.
- \(n\) is the sample size.

### Given:
- Sample mean (\(\bar{x}\)) = 75.9
- Population standard deviation (\(\sigma\)) = 10.6
- Sample size (\(n\)) = 90
- Confidence level = 99%

### Steps to Solve:

1. **Find the critical value \(Z_{\alpha/2}\) for a 99% confidence level.**
   - For a 99% confidence level, \(\alpha = 1 - 0.99 = 0.01\).
   - \(Z_{\alpha/2}\) corresponds to the value that leaves 0.5% in each tail of the normal distribution, which is approximately 2.576.

2. **Calculate the standard error (SE):**

\[
\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{10.6}{\sqrt{90}} \approx 1.117
\]

3. **Calculate the margin of error (ME):**

\[
\text{ME} = Z_{\alpha/2} \times \text{SE} = 2.576 \times 1.117 \approx 2.877
\]

4. **Calculate the confidence interval:**

\[
\text{CI} = 75.9 \pm 2.877
\]

- **Lower limit**: \(75.9 - 2.877 = 73.0\)
- **Upper limit**: \(75.9 + 2.877 = 78.8\)

### Final Confidence Interval:
The 99% confidence interval for the true mean of all scores is \( \textbf{[73.0, 78.8]} \).

Do you want details or have any questions?

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Here are 8 related questions you might find useful:

1. How is the critical value \(Z_{\alpha/2}\) determined for different confidence levels?
2. What happens to the confidence interval if the sample size increases?
3. How would the confidence interval change if the population standard deviation were unknown?
4. Why do we use the standard normal distribution for this calculation?
5. Can you calculate a 95% confidence interval for the same data?
6. What is the interpretation of a 99% confidence interval?
7. How is the margin of error affected by the sample size?
8. How would the interval change if we had a larger sample standard deviation?

**Tip:** Confidence intervals are used to estimate population parameters; the wider the interval, the more uncertainty in the estimate.

Learn how to calculate a 99% confidence interval for test scores using the sample mean, standard deviation, and critical values from the normal distribution. Detailed step-by-step solution with computations rounded to one decimal place.

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