Posted by Danielle on Thursday, January 24, 2008 at 5:55pm.
You are evidently studying bivariate distributions, and I am not an expert on that subject. Convert the 90th percentile to the number of standerd deviations from the mean score. I get 1.28 sigma. With a correlation of r=0.57, the number 1-r^2 = 0.675 probably plays a role in the number of standard deviations from the mean Math score. That would be 0.864 sigma, or the 80th percentile.
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