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Asked Oct 4, 2019
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Assume that scores on a bone density test are normally distributed with a mean of 0 and a deviation of 1. 

c. for a randomly selected subject, find the probability of a score between 0.87 and 1.78. 

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It follows normal distribution with mean 0 and standard deviation 1

for a randomly s...

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1.78) - P (Z < 0.87) = 0.962 - 0.808 = 0.154 P(Z NORM.S.DIST (1.78, TRUE)) P(Z 1.78) 0.962(From Excel 0.808(From Excel = NORM.S. DIST (0.87, TRUE)) P(Z < 0.87)

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