The combined SAT scores for the students at a local high school are normally distributed with a mean of 1531 and a standard deviation of 294. The local college includes a minimum score of 1619 in its admission requirements.

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

\[ P(X > 1619) = \, \% \]

Using Z-Scores and Normal Distribution table, we estimate that about 38.3% of students from the local high school would meet the SAT score requirement of the local college.

Explanation:

To answer this question, we need to convert the SAT score of interest, 1619, into a z-score and then determine the percentage of students whose scores fall above this z-score.

First, find the z-score for 1619: z = (X - μ) / σ, where X is the score of interest, μ is the mean, and σ is the standard deviation. For this problem, X = 1619, μ = 1531, and σ = 294. Plugging these values in, we get z = (1619 - 1531) / 294 ≈ 0.3.

Next, since we need the percentage of students who score above this z-score, we refer to the z-table which gives the cumulative probability up to a given z-score. For z = 0.3, the cumulative probability is approximately 0.617. However, since we need the area to the right of this z-score, we subtract this value from 1 (since the total area under the normal distribution curve is 1). So, 1 - 0.617 = 0.383 or 38.3%.

Therefore, approximately 38.3% of students from the high school are likely to have scores that satisfy the local college's admission requirement.

Learn more about Z-Scores and Normal Distribution here:

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