The combined SAT scores for the students at a local high school are normally distributed with a mean of 1526 and a standard deviation of 299. The local college includes a minimum score 1825 in its admission requirements. What percentage of students from this school earn scores that satisfy the admission requirement? Percentage % Please show your answer as a percent accurate to 2 decimal places (do not enter the %" sign).

About 16% of students from a school with a mean SAT score of 1526 and a standard deviation of 299 are likely to achieve the minimum SAT score of 1825 required for admission into the local college.

Explanation:

To answer this question, we will use the concept of the standard normal distribution. This is a special case of a normal distribution where the mean is 0 and standard deviation is 1. We will convert the given SAT score to a z-score, which represents the number of standard deviations it is away from the mean.

To calculate the z-score, subtract the mean from the given score and divide by the standard deviation. For an SAT score of 1825, with a mean of 1526 and a standard deviation of 299, the z-score is (1825-1526)/299 = 1.

To find the percentage of students earning scores that meet or exceed this z-score, we will look to the standard normal distribution table or use a calculator with a normal distribution feature. For a z-score of 1, the table indicates that approximately 16% of scores fall above this value, which translates to the percentage of students from this school that meet or exceed the required SAT score.

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