test2_f02sol

test2_f02sol - Solutions to ST 370 Online Second Exam, Fall...

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Solutions to ST 370 Online Second Exam, Fall 2002, Version A. 1. continuous random variable 2. [1-(1-.8)^3][1-(1-.9)^2]=.982 Since the two parallel circuits are in series, one containing 1,2,3 and the other containing 4 and 5, hence both have to work properly. So let us consider the first parallel subsystem containing 1,2,3: for this subsystem to work, it is the same as the complement of the event that this parallel circuit fails. Hence, (1-.8)^3 is the probability of all the three failing, but since we want the probability of the complement of this event, we subtract the probability from 1. A similar argument follows for the other parallel system. 3. A and B are independent events. This is essentially the definition of independence. 4. .9961-.0918=.9043 (from Table D) 5. 40(.67 or .68)+600=627: here, we find out the percentile for the standardized normal variate Z=(X-600)/40, and then getting .67 or .68, we get back the percentile in the X variate case by multiplying with the standard deviation and adding the mean. 6. [12!/(10!)(2!)][.75^10][.25^2]=.2323 (from the binomial pmf)
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test2_f02sol - Solutions to ST 370 Online Second Exam, Fall...

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