Question

Question is about z-scores....?

All students who complete Math III course at Little Hills High School take a common final exam. The exam scores are normally distributed with a mean of 105 and a standard deviation of 16.

a. Kyle and Ethan are Algebra 2 students who took the final exam. Kyle's score was 135 and Ethan's score was 93. Calculate the z-score for each student. Round your answers to the nearest tenth.

b. What percent of the students had a final exam score lower than Ethan's score?

I'm confused as to how to round my answer for part a....do I round the answer I got from using the formula to find the z-score? or do I round the answer I got from the z-score table?

Answer 1

a) Round the actual z-scores for Kyle (1.9) and Ethan (-0.8)
b) About 21.19%.

Explanation:

For part a, given that the question asked you to calculate the z-score for each student, I would say that the "Round your answers" portion refers to the z-scores. In Part a you are not doing anything beyond finding the z-scores.

Using the z-score formula `z = (x-mu)/sigma`:

Kyle: `z_K = (135-105)/16 = 30/16 = 1.875 ~~ 1.9`

Ethan: `z_E = (93-105)/16 = -12/16 = -0.75 ~~ -0.8`

Note: Ethan's z-score rounding assumes a rounding rule that ignores negative signs. There are quite a few rounding rules that can be followed for rounding negative values which don't affect the theory behind this solution, just the resulting percent in Part b [slightly].

To find the percent of students with a lower exam score than Ethan is essentially asking to find the proportion of area under the standard normal distribution `N(0,1^2)` curve for all z-scores < -0.8. This is known as a cumulative from negative infinity (aka "left-tail") area, which can be read from a z-score table by referencing a z-score of -0.8. That value comes out to roughly 0.2119.

In other words, about 21.19% of exam scores are lower than Ethan's. (It would be about 24.2% if you'd chosen to round towards 0 for Ethan's z-score.)