The combined SAT scores for the students at a local high school are normally distributed with a mean of 1514 and a standard deviation of 290. The local college requires a minimum score of 2123 for admission. What percentage of students from this school earn scores that satisfy the admission requirement?

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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