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September 19th, 2016, 05:19 PM   #1
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Average Absolute Value of the z-scores From The Normal Distribution Table

The Normal Distribution Table can tell you the probability of a value being more or less than a certain z-score. The Normal Distribution Table can also tell you for a given probability what the z-score would be. If the probabilities were divided into infinitely small units and the absolute value of the z-score at each probability was taken, what would the average absolute value of the z-scores be? I have data that I want to use the Normal Distribution Table for. Ignoring the fact that my data may or may not be symmetrically distributed, my data has a smaller standard deviation than normally distributed data. I'll give an example of what I want to do. Let's say I have symmetrical data where 80 percent of the data is within 1 standard deviation of the mean and 98 percent of the data is within 1 standard deviation of the mean. Is there a good transformation I can do with my data to change the standard deviations to make it normally distributed?

Edit: I generated 100 random probabilities from 0 to 1. I'm going to look up the z-score for each probability and use that to estimate the average absolute value of the z-scores for the Normal Distribution Table.

Last edited by EvanJ; September 19th, 2016 at 05:43 PM.
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