QUESTION
A project manager has been asked by the client to meet
the promise date of the project. The project manager analyzes the schedule
before promising a date to the customer. The project manager uses the program
evaluation and review technique to evaluate the project schedule. She decides
that based on the PERT calculations she can promise a delivery date of June 30.
The expected value of the project completion date is May 30. If the project
manager is willing to accept a 5% probability that the project will be delivered
later than June 30, what is the standard deviation of the duration of the
activities on the critical path? Assume a five-day workweek.
a. Ten days
b. Fifteen days
c. One-half
month
d. One month
Answer --- b
Most of the places I saw that this question is wrongly answered as option C =45 days(one and half month), without any justification.
I saw a blog in PMHUB, and many other sites , all are just copying , as 45 days without knowing the reasons why.
Some of them are little adding it with 95 % probability (appx) of 2sigma and thus of normal distribution but still they are answering 1.50month.
If It is by 2sigma of NDC, then one sigma = 31/2= 15days appx, also answer comes B. But it is wrong approach.
at here 95 % is different then
what
shown in NDC figure
see 47.87 (LHS)+47.87(RHS) = 95.73 %
where as in question it is
50(LHS) + 45 (RHS) = 95%
it makes a great difference
47.87-45 = 2.87%
with this difference at ends (flattered curve) = 45will give 1.65SIGMA, (value of Z, in Z score curve), while 47.87 will give = 2sigma.
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It can be checked in Excel program also
There is one formula NORMINV ( prob %age in decimal, MEAN DATE, SD) will give END date (pessimestic date).
But in this question given M = a Mean date, Prob %age = 95% and End date and asking SD =?
Could not be solved , directly.
But by indirect , by GOAL seek function of Excel =
Putting in formula NORMINV (0.95,30th May,SD) = 30th of June, by goal seek will give SD = 18.80 days( it also works on Normal Distribution Curve theory)
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BY PERT formula SD = P-O/6, Beta distribution , weighted average, method
Assuming -3sigma as Optimistic values = 3*31/1.645= 54 days, and pessimistic as cutoff date at 95% prob = 31 days from mean
= p-o/6 = 85/6 =14.1 appx = 15 days.
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By no other way answer will come.
