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natural units (X), standard deviations (Z), and percentiles (%)
Example of natural units--1150 on the SAT. Example of percentile: 55th percentile.
Other examples of natural units--inches, dollars, test score (note that getting 60 percent on a test is not the same as being in the 60th percentile!)
If the population distribution is normal, can go back and forth between natural units and percentiles, by converting to Z.
natural units to percentile
Example. What percentile is 1150 on SAT, if mean SAT is 1100 and standard deviation is 100? (those are made-up parameters)
percentile to natural units
Example. What score would you need on the SAT to get into the 85th percentile?
The case of the missing s. Suppose we were given the following information:
Naomi is in the 8th percentile for height. Naomi is 60 inches tall. The mean is 64 inches. What is the 20th percentile for height?
At first, this seems impossible to do, because we do not know s. However, in fact we can infer s from the fact that we are given the mean and one observation in both natural units and percentiles. Here is how:
Find Z for 8th percentile, using invnorm. invnorm (.08) = -1.45
Use algebra to solve for s. Z = (X - m)/s, so
-1.45 = (60-64)/s; s = -4/-1.45 = 2.76
Now that we know s, we can find the 20th percentile. We first find invnorm(.2) = -0.84, which is the Z that corresponds to the 20th percentile. To find X, we use
Z = (X - m)/s; -0.84 = (X - 64)/2.76; X = 61.69
Summary. Understand the relationship between natural units, Z, and percentiles. To get from natural units to percentiles, you use the Z transformation and normcdf. To go from percentiles to natural units, you use invnorm and the Z transformation. To solve the case of the missing s, you use a point where you are given both a percentile and an X value. You use the percentile and invnorm to find Z, then use the Z transformation and X to find s.