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Answer :
Final answer:
Using Z-scores and a normal distribution model, about 10.56% of students would be eligible for college A, and to be accepted into college B, a student would need a minimum SAT score of approximately 686.
Explanation:
Given that the SAT scores are normally distributed with a mean (μ) of 500 and a standard deviation (σ) of 80, we can apply the concept of Z-scores to find the percentage of students eligible for college A and the minimum score needed for acceptance into college B.
(a) College A accepts students with scores above 600. To find the percentage of students who score this high, we need to calculate a Z-score, which is (X - μ) / σ = (600 - 500) / 80 = 1.25. Consulting a Z-score (standard normal) table, the percentile rank for a Z-score of 1.25 is about 89.44%. That means that about 89.44% of students have scores of 600 or less, so College A would be open to approximately 100% - 89.44% = 10.56% of students.
(b) College B accepts the top 1% of scores. Again, consulting a Z-score table, the Z-score that corresponds to the top 1% (or the 99th percentile) is about 2.33. To convert this back into a score, we use the formula X = μ + (Z * σ) = 500 + (2.33 * 80) = 686.4. So a student would need a minimum SAT score of approximately 686 to be accepted to College B.
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