What percentage of students from this school earn scores that satisfy the admission requirement? P(X>917)=.....%

Question

The combined SAT scores for the students at a local high school are normally distributed with a mean of 1461 and a standard deviation of 302. The local college includes a minimum score of 917 in its admission requirements.What percentage of students from this school earn scores that satisfy the admission requirement? P(X>917)=.....%

Answer 1

Hi Natalie,

To find the percentage of students from the local high school who earn scores that satisfy the admission requirement (P(X > 917)), we need to use the z-score and the standard normal distribution table (or calculator).

First, we need to calculate the z-score for a score of 917 using the given mean and standard deviation:

Z = (X - μ) / σ

Z = (917 - 1461) / 302

Z = -544 / 302

Z ≈ -1.8013

Now, we look up the z-score of -1.8013 in a standard normal distribution table (or use a calculator) to find the percentage of scores greater than -1.8013. The percentage will be the percentage of students who meet the admission requirement.

Using a standard normal distribution table or calculator, you can find that P(Z > -1.8013) is approximately 0.9649.

To express this as a percentage, multiply by 100:

0.9649 * 100 ≈ 96.49%

So, approximately 96.49% of the students from the local high school earn scores that satisfy the admission requirement (scores greater than 917).

I hope this will help. I am happy to tutor you on any other questions you may have; please feel free to shoot me a message!