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ID: 404077 • Letter: #

Question

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Given a simple Project like the one pictured above. Optimistic / Most Likely / Pessimist Times were assessed and Path Means and Path Standard Deviations were calculated. For Path 1-2-5 we have a Path Mean = 31 and a Path Standard Deviation = 4. For Path 1-3-4-5 we have a Path Mean = 30 and a Path Standard Deviation = 2. Determine both the Probability that Path 1-2-5 will finish in 31 Days or Less and the Probability that the Project will finish in 31 Days or Less. 0.977249938 : 0.977246615 0.5 : 0.345731234 0.77337272 : 0.755778443 0.987775567 : 0.987775283 0.977249938 : 0.977246615 0.5 : 0.345731234 0.77337272 : 0.755778443 0.987775567 : 0.987775283 0.977249938 : 0.977246615 0.5 : 0.345731234 0.77337272 : 0.755778443 0.987775567 : 0.987775283 Given a simple Project like the one pictured above. Optimistic / Most Likely / Pessimist times were assessed and Path Means and Path Standard Deviations were calculated. For Path 1-2-5 we have a Path Mean = 31 and a Path Standard Deviation = 4. For Path 1-3-4-5 we have a Path Mean = 30 and a Path Standard Deviation = 2. Determine both the Probability that Path 1-2-5 will finish in 31 Days or Less and the Probability that the Project will finish in 31 Days or Less. 0.977249938 : 0.977246615 0.5 : 0.345731234 0.77337272 : 0.755778443 0.987775567 : 0.987775283

Explanation / Answer

Let Z be a normal random variable with mean mu and std. dev. of sigma. Consider a transformation: X = (Z - mu)/sigma. Then X is a standard normal random variable with a mean of 0 and std. deviation of 1.


So, for path 1-2-5 : P(Z < 31) = P(X < (31 - 31) / 4) = P(X < 0) = 0.5

P(Project completed in 31 days) = P(path 1-2-5 finishes in 31 days) + P(1-3-4-5 finishes in 31 days) - P(both finish in 31 days). Like previous case: P(1-3-4-5 finishes in 31 days) = 0.6915. So,

P(Project completed in 31 days) = 0.8457


So, by narrowing down the options only second one is correct.


NOTE: It is not clear that what all paths are needed to complete the project, so the second answer is tentative