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Normal quantiles
Posted:
Sep 25, 1996 11:03 AM


Here is a discussion of normal quantiles written by Jeff Witmer of Oberlin College. I think it sheds considerable light on the situation. As you see in the memo from Nancy Acree, Minitab uses a slightly different approximation, but it is very much in the same spirit.
Dick Scheaffer
 Begin Included Message  Reply to: RE>QQ
A useful way to think about the kth normal score is to think of the expected value of kthlargest observation in a sample of size n from a Normal(0,1) distribution. For example, if n=3 then the 1st normal score is the expected value of the minimum in a sample of size 3 from a standard normal. (Note that this is _not_ .68, which is the 25th percentile (number that cuts off the bottom 25%) of a Z distribution.)
If we were sampling from a uniform distribution on (0,1), then we would expect the 3 observations, on average, to divide the interval into four subintervals of length .25 each. Thus, we would expect the minimum to be .25. If you do a simulation this is in fact what happens. For example, I had my computer generate 1000 random samples of size 3 from a uniform(0,1) distribution, then I found the minimum in each sample; the average of these 1000 minimums was 0.246.
When sampling from a normal, however, the expected value of the minimum is not the 25th percentile (.68). Rather, the expected value is around .9. (I did a simulation of normal data on my computer and came up with an average of .908.) Moreover, the sampling distribution of the minimum (from a sample of size 3 from a standard normal) is skewed to the left. Thus, some people (i.e., some software developers) use the median, rather than the mean, of this sampling distribution. The median of the sampling distribution of the minimum of 3 observations from a Normal(0,1) is around .85. This is the value of the first normal score, when n=3, that is produced by some software. (In the simulation I ran on my computer, the median was .849.)
The general version of this is that the kth normal score for a sample ofsize n is given by the inverse Gaussian cdf at (k1/3)/(n+1/3). (When n=3 and k=1, this is the inverse Gaussian cdf at (2/3)/(10/3), or the .20 percentile, which is around .85.)
Jeff 



