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Answer :
Final answer:
Approximately 4.8% of students from the local high school earn scores that satisfy the admission requirement.
Explanation:
To find the percentage of students from the local high school who earn scores that satisfy the admission requirement, we need to convert the minimum admission requirement score to a z-score.
First, we calculate the z-score using the formula:
z = (x - mean) / standard deviation
where x is the minimum admission requirement score, mean is the mean SAT score of the local high school (1514), and standard deviation is the standard deviation of the SAT scores (290).
Substituting the values, we get:
z = (2123 - 1514) / 290
Simplifying, we find:
z ≈ 1.6655
Next, we need to find the percentage of students who have a z-score greater than or equal to 1.6655. We can use a standard normal distribution table or a calculator to find this percentage.
Using a standard normal distribution table, we find that the percentage of students with a z-score greater than or equal to 1.6655 is approximately 4.8%.
Therefore, approximately 4.8% of students from the local high school earn scores that satisfy the admission requirement.
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