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This problem involves determining the required sample size to estimate the mean SAT score with a given margin of error, confidence level, and standard deviation. 

To find the required sample size \( n \), we use the formula for margin of error in estimating a population mean:

\[
E = z \frac{\sigma}{\sqrt{n}}
\]

Where:
- \( E \) is the margin of error (given as 25),
- \( z \) is the z-score corresponding to the 95% confidence level (which is 1.96 for a two-tailed test),
- \( \sigma \) is the population standard deviation (given as 300),
- \( n \) is the sample size (which we need to find).

Rearranging the formula to solve for \( n \):

\[
n = \left( \frac{z \sigma}{E} \right)^2
\]

Substituting the known values:

\[
n = \left( \frac{1.96 \times 300}{25} \right)^2
\]

Let me calculate this for you.The required sample size is 553 students to achieve a margin of error of 25 points with a 95% confidence level.

Would you like further details or have any questions?

Here are five related questions to explore further:
1. How does increasing the confidence level affect the required sample size?
2. What if the margin of error were reduced to 20 points instead of 25?
3. How would a smaller standard deviation impact the sample size?
4. How is the z-score determined for different confidence levels?
5. What are the assumptions behind using this formula for sample size estimation?

**Tip:** Remember, increasing the sample size reduces the margin of error, making the estimate more precise!

SAT scores are distributed with a mean of 1,500 and a standard deviation of 300. You are interested in estimating the average SAT score of first-year students at your college. If you would like to limit the margin of error of your 95% confidence interval to 25 points, how many students should you sample?

This problem involves determining the required sample size to estimate the average SAT score with a margin of error of 25 points at a 95% confidence level. Learn how to apply the formula for margin of error and confidence intervals with a standard deviation of 300 and a mean SAT score of 1,500.

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