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OpenStudy (anonymous):
X has a normal distribution, and P(X>73.05)=0.0289. Given that the variance of the distribution is 18, find the mean.
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OpenStudy (dumbcow):
same equations Z = (73.05 - mean)/sqrt18 when you look up the probability, you need to do 1-.0289, because of how they set up the probability tables
OpenStudy (amistre64):
on some tables it measures from themean itself ..... so .5 - .0289 might be it as well
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