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To solve this problem, we need to calculate the sample size required to estimate the average SAT score with a margin of error of 10 points for a 92% confidence interval.

### Step 1: Identify the given data.
- Population standard deviation (\(\sigma\)) = 300
- Margin of error (E) = 10
- Confidence level = 92%

### Step 2: Determine the Z-value for a 92% confidence level.
We can find the Z-value for a 92% confidence interval using a Z-table or statistical software. The confidence level of 92% corresponds to a Z-value of approximately 1.75.

### Step 3: Use the formula for margin of error:
The formula for margin of error in a confidence interval is:

\[
E = Z \cdot \frac{\sigma}{\sqrt{n}}
\]

Where:
- \(E\) is the margin of error
- \(Z\) is the Z-value corresponding to the confidence level
- \(\sigma\) is the population standard deviation
- \(n\) is the sample size

### Step 4: Rearrange the formula to solve for \(n\) (sample size):
\[
n = \left( \frac{Z \cdot \sigma}{E} \right)^2
\]

### Step 5: Plug in the values:
\[
n = \left( \frac{1.75 \cdot 300}{10} \right)^2
\]

\[
n = \left( \frac{525}{10} \right)^2 = (52.5)^2 = 2756.25
\]

Since we need a whole number, we round up to the next whole number:

\[
n = 2757
\]

### Conclusion:
The administrator should sample **2,757** students to achieve a margin of error of 10 points with a 92% confidence interval.

Would you like any more details or have any questions?

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Here are some related questions for further understanding:
1. How does increasing the confidence level affect the required sample size?
2. What happens to the sample size if the margin of error is reduced to 5 points?
3. Why is the sample size rounded up instead of down?
4. What role does the standard deviation play in determining the sample size?
5. How would the sample size change if the population standard deviation was estimated instead of known?

**Tip:** Always round the sample size *up* when using this method to ensure the desired margin of error is achieved!

SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300. An administrator at a college is interested in estimating the average SAT score of first-year students. If the administrator would like to limit the margin of error of the 92% confidence interval to 10 points, how many students should the administrator sample?

Estimate the sample size needed to limit the margin of error for estimating SAT scores with a 92% confidence interval. The problem uses a normal distribution with a mean of 1,500 and a standard deviation of 300. Learn to apply the margin of error formula to find the required sample size.

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