How to Find a Percentile for a Normal Distribution
A popular normal distribution problem involves finding percentiles for X. That is, you are given the percentage or statistical probability of being at or below a certain xvalue, and you have to find the xvalue that corresponds to it. For example, if you know that the people whose golf scores were in the lowest 10% got to go to a tournament, you may wonder what the cutoff score was; that score would represent the 10th percentile.
A percentile isn’t a percent. A percent is a number between 0 and 100; a percentile is a value of X (a height, an IQ, a test score, and so on).
Certain percentiles are so popular that they have their own names and their own notation. The three “named” percentiles are Q_{1} — the first quartile, or the 25th percentile; Q_{2} — the 2nd quartile (also known as the median or the 50th percentile); and Q_{3} — the 3rd quartile or the 75th percentile.
Here are the steps for finding any percentile for a normal distribution X:

1a.If you’re given the probability (percent) less than x and you need to find x, you translate this as: Find a where p(X < a) = p (and p is the given probability).
That is, find the pth^{ }percentile for X. Go to Step 2.

1b.If you’re given the probability (percent) greater than x and you need to find x, you translate this as: Find b where p(X > b) = p (and p is given).
Rewrite this as a percentile (lessthan) problem: Find b where p(X < b) = 1 – p. This means find the (1 – p)th percentile for X.

Find the corresponding percentile for Z by looking in the body of the Ztable (see below) and finding the probability that is closest to p (from Step 1a) or 1 – p (from Step 1b).
Find the row and column this probability is in (using the table backwards). This is the desired zvalue.

Change the zvalue back into an xvalue (original units) by using
You’ve (finally!) found the desired percentile for X. The formula in this step is just a rewriting of the zformula,
so it’s solved for x.
Here’s an example: Suppose that you enter a fishing contest. The contest takes place in a pond where the fish lengths have a normal distribution with mean 16 inches and standard deviation 4 inches. Now suppose you want to know what length marks the bottom 10 percent of all the fish lengths in the pond. What percentile are you looking for?
Being at the bottom 10 percent means you have a “lessthan” probability that’s equal to 10 percent, and you are at the 10th percentile.
Now go to Step 1a and translate the problem. In this case, because you’re dealing with a “lessthan” situation, you want to find x such that p(X < x) = 0.10. This represents the 10th percentile for X. The following figure shows a picture of this situation.
Now go to Step 2, which says to find the 10th percentile for Z.
Looking in the body of the Ztable, the probability closest to 0.10 is 0.1003, which falls in the row for z = –1.2 and the column for 0.08. That means the 10th percentile for Z is –1.28; so a fish whose length is 1.28 standard deviations below the mean marks the bottom 10 percent of all fish lengths in the pond.
But exactly how long is that fish, in inches? In Step 3, you change the zvalue back to an xvalue (fish length in inches) using the zformula solved for x; you get x = 16 + –1.28[4] = 10.88 inches. So 10.88 inches marks the lowest 10 percent of fish lengths. Ten percent of the fish are shorter than that.