# What is the best way to interpolate over the 25th and 75th percentile of SAT scores?

The problem:

I know the 25th and 75th percentiles of SAT scores for students admitted to a given university, and I want to interpolate over those two points in order to estimate all the percentiles (i.e. 1st-100th) of scores for students admitted to the university.

What I know about SAT score distributions:

1. SAT scores must be in the interval [600, 2400] and are approximately normally distributed on a nationwide basis: 1. Some extremely competitive universities, such as MIT and Harvard, may have the highest possible SAT score (i.e. a 2400) at their 75th percentile, so I'm guessing their distribution might be truncated on the right side (not sure if this would still be a normal distribution?).

2. I have a histogram of all 1,547,990 SAT scores taken in 2010 including the mean and standard deviation: http://professionals.collegeboard.com/profdownload/sat-percentile-ranks-composite-cr-m-w-2010.pdf.

• Can you be more specific. I am having trouble understanding your question. do you want to reproduce a normal distribution given the 25th and/or 75th percentiles? – mandata Jun 15 '15 at 20:02
• I essentially want to estimate the percentiles other than the 25th and 75th. For example, I already know the 25th and 75th percentile for SAT scores for students admitted to MIT, but I also want to estimate all the other percentiles, such as the 29th, 56th, and 83rd percentiles. – Brian Schmitz Jun 15 '15 at 20:09
• you have to use a normal table to get the z-score. with the z-score, you can calculate the sat score. see. en.wikipedia.org/wiki/Standard_score. – mandata Jun 15 '15 at 20:15

Your post is a bit confusing--to me, at least. The original question alone is interesting. Along with just the assumption that scores are normal, it contains enough information to give a good answer. (It is not clear whether items 1-3 are part of the question, or your own research towards an answer. They are also worthwhile information for many purposes, but I will ignore them here--except for normality.)

(a) The mean of a normal distribution is halfway between the 25th and 75th peercentiles (also called lower and upper quartiles). Average these two values to approximate the population mean.

(b) In a normal distribution, the difference between these two percentiles is about 1.35 times its standard deviation. So take the difference between these two values and divide by 1.35 to approximate the population standard deviation.

You can verify both (a) and (b) by looking at printed standard normal tables. Also, you might be able use selected data from your items 1-3 to see that both (a) and (b) work well in practice.

The parameters of the normal family are the mean and standard deviation (SD). Once they are known, the distribution is completely specified. For example, knowing the mean and SD, you could find the proportion of students in the population that score above 2000.

When administered to a large and diverse population, scores of most standardized tests tend to be normal (roughly) because of the Central Limit Theorem. Much of the interpretation of college-entrance exam scores depends on normal distribution and published information about means and various percentiles. Publishers of these exams can fine tune questions and scoring so that the distribution of scores for any one test are even more closely normal than would necessarily be so on theoretical grounds.

• I may lack some statistical knowledge, but this is why I included points 1-3. (1) I just wanted to show that applicant scores are roughly normally distributed. (2) I wasn't sure if the distribution of admitted applicants at competitive universities would still be normal if the right side of the distribution was truncated, i.e., since a large number of strong applicants would have the exact same, maximum possible score (2400). (3) I didn't know if any additional data would be needed, so I just included all the data. – Brian Schmitz Jun 15 '15 at 20:29
• @BrianSchmitz: your point (2) is a very good one. There is no reason to suspect that the admitted students have a normal distribution. For example, a university might just admit all the highest scores among its applicants, choosing the cutoff to get the right number. This could lead to a distribution similar to the one above, but cut off with nothing below 1900, for example. The 25th percentile of admits might be 1950 and the 75th percentile might be 2200 (to my eye). Of course, the applicant pool may not be normally distributed either. – Ross Millikan Jun 15 '15 at 20:52
• That's a good point. For simplicity, let's assume a model where there are no score lower cutoffs, where the applicant pool is normally distributed, and where we don't know the distribution of admitted applicants. – Brian Schmitz Jun 15 '15 at 21:15
• I agree that's what likely what was intended. This is what happens when the student brings more careful thought to the practicality of a problem than may have been intended by the authors. (Also, a fundamental flaw in multiple-choice tests which indoctrinate students to believe there is only one correct answer to a question.) – BruceET Jun 15 '15 at 21:28
• Without actual data on the distribution of scores for accepted students, I don't see how to approach a more accurate solution. – BruceET Jun 15 '15 at 21:58