The combined SAT scores for the students at a local high school are normally distributed with a mean of 1504 and a standard deviation of 304. The local college includes a minimum score of 2385.6 in its admission requirements.

What percentage of students from this school earn scores that satisfy the admission requirement?

To determine the percentage of students from the local high school who satisfy the admission requirement of a local college, we need to find the percentage of students whose SAT scores are equal to or higher than the minimum score of 2385.6.

First, we need to calculate the z-score using the formula: z = (x - μ) / σ, where x is the SAT score, μ is the mean, and σ is the standard deviation.

Once we have the z-score, we can use a standard normal distribution table or calculator to find the percentage of students with z-scores equal to or greater than the calculated z-score.

Let's assume a student's SAT score is 2385.6. Using the formula, we calculate the z-score: z = (2385.6 - 1504) / 304 = 2.902.

When we look up the z-score in the standard normal distribution table or calculator, we find that approximately 99.71% of the students have SAT scores less than or equal to 2385.6.

Therefore, the percentage of students from the local high school who satisfy the admission requirement is approximately 99.71%.

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