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The combined SAT scores for the students at a local high school are normally distributed with a mean of 1512 and a standard deviation of 296. The local college includes a minimum score of 1453 in its admission requirements.

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

P(X> 1453) =___%

Enter your answer as a percent accurate to 1 decimal place (do not enter the "%" sign). Answers obtained using exact z

scores or z-scores rounded to 3 decimal places are accepted.

Answer :

Final answer:

Using z-scores and the properties of the normal distribution, we can find that approximately 57.9% of students from a given high school would satisfy the minimum SAT score admission requirement of a local college.

Explanation:

The goal is to find the percentage of students whose SAT scores are higher than 1453. This is a problem about the normal distribution, and we need to use the z-score formula: z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation.

First, we calculate the z-score for the score of 1453. z = (1453 - 1512) / 296 = -0.199. A negative z-score indicates that the score of 1453 is below the mean score of 1512. However, we are not interested in the z-score itself, but rather the area to the right of it, as this corresponds to the percentage of students who score higher than 1453.

We can find this area (the percentage of students) by using a z-table or online z-score calculator. Looking up the z-score of -0.199, we find that the area to the left is roughly 0.4215 or 42.15%. But we need the area to the right, so we subtract this from 1 (or 100%): 1 - 0.4215 = 0.5785 or 57.85%.

So, approximately 57.9% of students from the school will earn SAT scores that satisfy the admission requirement of the local college.

Learn more about Normal Distribution here:

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