Score  Proportion of Students 
95100  .01 
9095  .02 
8590  .03 
8085  .08 
7580  .21 
7075  .15 
6570  .1 
6065  .2 
5560  .1 
5055  .1 
For each range of test scores, a certain proportion of students achieved that score. What your instructors are doing when they list test scores like this is providing you with the distribution of test scores. Notice that if you add up all of the proportions they add up to 1 (i.e. 100%).
You can also illustrate a distribution graphically if you want.
So your instructor may show the grades like this instead:
In fact when you encounter discussions of distributions in the psychological
literature they are typically depicted graphically. The important
point is that all a distribution is telling you is the relative frequency
of various scores.
Now remember that a distribution is simply a collection of scores. Because of that, like any other collection of scores you can calculate descriptive statistics for those scores.
One of the easiest approaches to dealing with this problem is to use the standard deviation as your yardstick. If you know the mean GPA at each university and you know the standard deviation of the GPAs at each university then you can provide each applicant with an adjusted score that indicates how many standard deviations above or below the mean each applicant is.
Say Fred has a mean GPA of 3.5. He comes from a University where the average GPA is 2.0 and the standard deviation is .5. Fred's GPA is three standard deviations above the mean. Say, Beevis has a GPA of 3.5. He comes from a University where the average GPA is 3.0 and the standard deviation is .25. Beevis's GPA is two standard deviations above the mean. Assuming the overall quality of the students at both universities is the same (which could be checked with incoming SATs and so on) then it appears that Fred is the better applicant, even though they both have the same GPA.
Statisticians use the term Z score to refer to how many standard deviations above or below the mean a score is. If someone's Z score equals 3, that means their score is three standard deviations higher than the average score. If someone has a Z score of 2, that means their score is two standard deviations below the average score. And so on.
Z scores are important because they allow you to take things that are measured on different scales an equate them. Consider for instance, the SAT and ACT. The SATs were designed to have means of about 500 for each subscale and standard deviations of about 100 (these have changed over time with the population, but this was true of the original test). ACTs as you know have a much lower mean and standard deviation. Imagine you wanted to compare two students, one who took the SAT and one who took the ACT. How would you do it? The answer is really pretty simple, calculate the Z score for each student. The Z score puts both tests on the same scale.
Notice that this is plotted in terms of Z scores. You can also plot one in terms of raw scores. If you do so, it will maintain the same basic shape except that the X Axis will be either stretched or shrunk. In fact, imagine that this is printed on a big rubber sheet and that it can be stretched or shrunk along the X Axis as much as you like.
Before you go too far in this tutorial remind yourself of what all this means. Z Scores are simply how many standard deviations from the mean a score is. So when we say that 84% or so of scores fall below a Z score of 1, what we're really saying is that if your score is one standard deviation above the mean, then 84% of the scores are below you. Or in other words when you're one standard deviation above the mean, your percentile rank is 84.
Z Score Scores Below Scores Above 3 0.001349967 0.998650033 2 0.022750062 0.977249938 1 0.15865526 0.84134474 0 .5000000000 .500000000 1 0.84134474 0.15865526 2 0.977249938 0.022750062 3 0.998650033 0.001349967
You can find Z scores in the tables at the end of almost every stats book and many research methods books. If you're familiar with excel there's also an easy function you can use to calculate the proportion of scores that fall below a particular Z score. The function is called NORMSDIST. Say you wanted to know how many scores fall below a Z score of 1.5. You would type the following into a cell in your excel worksheet:
When I did that the answer I got was 0.933192771. So about 93% of all scores fall below a Z score of 1.5. To calculate the proportion of scores above that Z score you would subtract that answer from 1. So you could type in:
=normsdist(1.5)
The answer I got for that was 0.066807229. So one of the most useful things about a normal distribution is that once you know a Z score you know the proportion of scores that fall above or below that score.
=1normsdist(1.5)
Your problem is that you want to accept about 15% of the applications and you want to know, where you should put the cutoff such that 15% of the applicants will be accepted and 85% will be rejected. This problem then, is just the reverse of the problem discussed above. For what Z score, will 15% of the scores be above that Z score and 85% be below that Z score.
The answer to this problem is quite easy. If knowing a Z score
tells you the proportion of scores below that score, then the reverse must
also be true. If you know what proportion of scores fall above or
below a score, you can derive the Z score. Its just the inverse
operation. Again you can use the tables in any stats book.
You can also use a formula in excel called NORMINV. To use it you
need to type in the probability below the Z score and then 0 and
1 for the mean and standard deviation (when you convert to Z scores the
mean becomes 0 and the standard deviation becomes 1). Here's how
you would use the function if you wanted to know the Z score where 95%
of the scores fall below that score:
=NORMINV(.95,0,1) 
Notice the number I typed in was the proportion .95 not the percentage 95. It will only work if you use the proportion. The answer I got from excel was 1.644853. So if someone's percentile rank is 95, then their score is 1.64 standard deviations above the mean (i.e. they have a Z score of 1.64).
This probably goes without saying, but you can also solve the problem
if instead you're given the proportion of scores above a particular
cutoff. So say you want 8% of the score to fall above the cutoff.
You would now enter:
=NORMINV(1.08,0,1) 
When I did that the Z score I got was 1.405073817.
 If you know how many standard deviations above or below the mean a score is (i.e. the Z score), you know what proportion of scores fall above or below that score.
 If you know the proportion of scores that fall above or below a score, you know how many standard deviations from the mean that score is.





