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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=(X600)/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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 Spring '08
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