Step 1 :The problem is asking for the sample size needed to estimate the average SAT score of first year students at a college with a margin of error of 25 points and a 95% confidence level.
Step 2 :The formula for the margin of error in a confidence interval is: \(E = Z * (\sigma/\sqrt{n})\), where \(E\) is the margin of error, \(Z\) is the Z-score (which corresponds to the desired confidence level), \(\sigma\) is the standard deviation, and \(n\) is the sample size.
Step 3 :We can rearrange this formula to solve for \(n\): \(n = (Z * \sigma / E)^2\)
Step 4 :We know that the standard deviation (\(\sigma\)) is 300, the margin of error (\(E\)) is 25, and the Z-score for a 95% confidence level is approximately 1.96.
Step 5 :We can substitute these values into the formula to find the required sample size: \(n = (1.96 * 300 / 25)^2\)
Step 6 :Calculating the above expression, we find that \(n = 554\)
Step 7 :Final Answer: The required sample size to limit the margin of error of the 95% confidence interval to 25 points is \(\boxed{554}\).