The combined math and verbal scores for students taking a national standardized examination for college admission, is Normally distributed with a mean of and a standard deviation of . If a college requires a minimum score of for admission, what percentage of students do not satisfy that requirement?

96.1%

Step-by-step explanation:

This is a normal distribution problem

μ = mean = 500

σ = standard deviation = 170

To solve this question, we require the normalized/standard/z-score value of 800.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (800 - 500)/170 = 1.765

To determine the percentage of student do not satisfy that requirement, this refers to students that do not score up to the minimum requirement of 800.

P(x < 800) = P(z < 800)

We'll use data from the normal probability table for these probabilities

P(x < 800) = P(z < 1.765) = 1 - P(z ≥ 1.765) = 1 - P(z ≤ -1.765) = 1 - 0.039 = 0.961.

This points to the fact that 96.1% of the candidates do not normally reach the minimum requirement.

Pls delete this answer. I answered incorrectly, and it will take a lot of time to re-answer it

(Btw, this is edited. I had the right answer, and then I realized that it's wrong)