a. Ten days
b. Fifteen days
c. One-half month
d. One month
what should be the answer and why? thanks in advance!
probability of completing on or before June 30 = 95% confidence level
probability of completing after June 30 = 5% confidence level
To get 95% probability of a date less than June 30, you will need to add 50% of the area on the left side of the bell (normal) curve and 45% of the area on the right side of the bell curve.
PERT = [O+4M+P] / 6
O = 30 days * 95% = 28.5 days
M = 30 day * 50% = 15 days
P = 30 days * 5% = 1.5 day
So PERT = [ 28.5 + (4*15) +1.5 ] / 6 = 90 / 6 = 15 days
So correct answer is C - 1/2 month
Again it does not say how much delay from 30th June. 5% is probability and not % delay. So may be we cannot say that since there is 5% probability of delay, there is 95% probability of early delivery. We don't know how much 'early' or 'late'. So we do not have optimistic and pessimistic estimate.
I did not get the crux of the question itself. May be I have to revisit time estimation T&T.
Can you please explain it in more detail?
After solving even I was debating between picking 15 days vs. 1/2 months but ended up with latter one.. I would not worry about such questions but would concentrate on pmp exam preparation and be thorough with everything as much as possible..
In order for the project to have a 5% probability of being late, there is a 95% probability that the project will be delivered on time or earlier.
I found this statistical formula -
Code: Select all
We can determine the Probability associated with a Range. This Probability is also called the Confidence Level. The Statistical Distribution Theory provides Probability for different values of ‘n’.
If n=1 then Probability is 68.27%
If n=2 then Probability is 95.45%
If n=3 then Probability is 99.73%
If n=6 then Probability is 99.999999%
In terms of the PERT calculation this means that 2 standard deviations should be added to the expected value date of May 30. Since there is a one month difference between promise date of June 30 and the expected value of May 30, we can take optimistic day = 0 (first day - 30 May), pessimistic day = 30 (end day - 30 June).
So one standard deviation = (30-1)/6 = 4.83 i.e., 5 days
Based on above table, n = 2 should be taken.
So standard deviation is 2* 5 = 10 days (second sigma), which is one-half month.
This is a better answer than fifteen days because on the basis of a five-day work week this is close to three weeks.
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