The combined SAT scores for the students at a local high school are normally distributed with a mean of 1550 and a standard deviation of 299. The local college includes a minimum score of 832 in its admission requirements. What percentage of students from this school earn scores that fail to satisfy the admission requirement?

\[ P(X < 832) = \]

To find the percentage of students who fail to satisfy the admission requirement, calculate the z-score for the minimum score of 832 and use the standard normal distribution to find the corresponding probability.

Explanation:

To find the percentage of students who earn scores that fail to satisfy the admission requirement, we need to find the probability that a student's SAT score is less than 832. We can use the standard normal distribution to solve this. First, we calculate the z-score using the formula:

z = (X - μ) / σ

where X is the value (832), μ is the mean (1550), and σ is the standard deviation (299). Substituting the values, we get:

z = (832 - 1550) / 299 ≈ -2.18

Next, we can use a standard normal distribution table or a calculator to find the probability corresponding to the z-score of -2.18. The percentage of students earning scores below 832 is approximately 1.87%. Therefore, about 1.87% of students from this school fail to satisfy the admission requirement.

Learn more about SAT here:

https://brainly.com/question/13497310

#SPJ11