At one university, the students are given z-scores at the end of each semester instead of traditional GPA's. The mean and standard deviation of all the student' culmulative GPA's, on which the z-scores are based, are 2.7 and .5 respectively.

i understand how to translate the pas given the z-scores; however, i don't understand this question:

the president of the university wishes to graduate the top 16% of students with cum laude honors and the top 2.5% with summa cum laude honors. where should the limits be set in terms of z-scores [approx]? in terms of GPAs? what assumption if any did you make about the distribution of the GPAs at the university?

am i supposed to assume that there's a normal distribution? i don't understand how to get the answers to this question.
thank you.

oh wait...should use the 97.5 percentile to get the answer to the 2.5% top students?? because they wouldn't have negative z-scores, right?

Yes, assume a normal distribution.

100 - 2.5 = 97.5%
so any F(z) > .975 gets Summa
My table of z versus F(z) is pretty crude.
for example it has entries
z = 1.9 when F(z) = .971
z = 2.0 when F(z) = .977
We know that somewhere between z = 1.9 and z = 2.0, F (z) = .975
Say maybe any z over 1.95 gets summa.
Now do the same thing for F(z) = 1-.16 = .84
find z for f(z) = .84 (z around 1.0)
any z between there and 1.95 gets cum laude

Yes

To answer this question, we need to make an assumption about the distribution of GPAs at the university. Based on the context and the use of z-scores, it is reasonable to assume that the distribution of GPAs follows a normal distribution.

To find the z-scores corresponding to the top 16% and top 2.5%, we can use the standard normal distribution table (also known as the z-table). The z-table provides the percentage of data below a given z-score.

1. Finding the z-scores for the top 16%:
Since we are interested in the top percentage, we need to find the z-score that corresponds to the area under the normal curve equal to 1 - 16% = 84%. We look for this value in the z-table, and we find it to be approximately 0.994.

2. Finding the z-scores for the top 2.5%:
Similarly, we need to find the z-score corresponding to the area under the normal curve equal to 1 - 2.5% = 97.5%. Using the z-table, we find this value to be approximately 1.96.

Now, we can find the corresponding GPAs associated with these z-scores. We will use the formula:

GPA = (z-score * standard deviation) + mean

Using the given mean of 2.7 and standard deviation of 0.5 and substituting the z-scores we found:

1. For the top 16%:
GPA = (0.994 * 0.5) + 2.7 ≈ 3.197

2. For the top 2.5%:
GPA = (1.96 * 0.5) + 2.7 ≈ 3.68

Therefore, to graduate cum laude, the GPA limit should be set around 3.197, whereas to graduate summa cum laude, the GPA limit should be set around 3.68.

Keep in mind that the assumption made about the distribution of GPAs being normal is an approximation, and the actual distribution may deviate from the normal distribution to some extent.