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Assume that 1,000 scores are normally distributed with a mean of 72 and a standard deviation of 4. How many scores can we expect to be between 75 and 82?
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z-score for 75 = (75-72)/4 = 3/4 = .75 z-score for 82 = (82-72)/4 = 10/4 = 2.5
use a table such as http://training.ce.washington.edu/pgi/Modules/08_specifications_qa/Images/main_pictures/normal_table.gif to find the area between them
actually that table sucks
P( .75 < Z < 2.5) = P(Z < 2.5) - P(Z < .75) = .9938 - .7734 = .2204
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thansk man
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