The combined SAT scores for the students at a local high school are normally distributed with a mean of 1520 and a standard deviation of 295. The local college includes a minimum score of 901 in its admission requirements.

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

\[ P(X > 901) = \]

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.

Question

23-09-2023
Mathematics

High School

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The combined SAT scores for the students at a local high school are normally distributed with a mean of 1520 and a standard deviation of 295. The local college includes a minimum score of 901 in its admission requirements.

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

\[ P(X > 901) = \]

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 :

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sheenasharma sheenasharma · Answer 1 · 2024-01-31T04:22:08-05:00

Final answer:

98.2% of students from this high school have SAT scores that meet the local college's admission requirement. This was calculated by finding the Z-Score of the minimum admission score, then referring to the standard normal distribution table to find the corresponding percentage.

Explanation:

In order to find the percentage of students who meet the admission requirements, you first need to find the Z-Score. The Z-Score can be calculated using the formula (X - μ) / σ where X is the score we are considering (901), μ is the mean (1520), and σ stands for standard deviation (295). So, the Z value will be (901 - 1520) / 295 = -2.09.

Next, we refer to the standard normal distribution table to find the probability associated with this Z-Score. Since we are looking for a value greater than -2.09, we need to subtract the value found on the table (0.0182) from 1 (because the total probability under the curve is 1). Thus, 1 - 0.0182 = 0.9818. Therefore, 98.2% of the students at the school satisfy the admission requirement of the local college.

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