Standard Deviation

Can somebody explain the following question please?


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

 


As per 3point estimate (PERT - wieghted average beta distribution method)


ans will be 15 days


refer earlier Question


http://www.pmzilla.com/mock-questions


http://er-sspawar.blogspot.in/

Thanks a lot.


 


Regards,


MK

THIS ARTICLE MAY HELP A LOT -from -

http://www.interventions.org/pertcpm.html

5.2. Probability of Project Completion by Due Date

Now, although the project is estimated to be completed within 28 weeks (te=28) our Project Director would like to know what is the probability that the project might be completed within 25 weeks (i.e. Due Date or D=25).

For this calculation, we use the formula for calculating Z, the number of standard deviations that D is away from te.

By looking at the following extract from a standard normal table, we see that the probability associated with a Z of -0.6 is 0.274. This means that the chance of the project being completed within 25 weeks, instead of the expected 28 weeks is about 2 out of 7. Not very encouraging.



 

 

On the other hand, the probability that the project will be completed within 33 weeks is calculated as follows:

The probability associated with Z= +1 is 0.84134. This is a strong probability, and indicates that the odds are 16 to 3 that the project will be completed by the due date.



 

 

If the probability of an event is p, the odds for its occurrence are a to b, where:

Select Bibliography

Wiest, Jerome D., and Levy, Ferdinand K., A Management Guide to PERT/CPM, New Delhi: Prentice-Hall of India Private Limited, 1974

Render, Barry and Stair Jr., Ralph M. - Quantitative Analysis for Management, Massachusetts: Allyn & Bacon Inc., 1982, pp. 525-563

Freund, John E., Modern Elementary Statistics, New Delhi: Prentice-Hall of India Private Limited, 1979

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.

-------------------------------------------------------------------------------------------------------

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)

-----------------------------------------------------------------------------------------------------------------------------------------

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.

--------------------------------------------------------------------------------------------------------------------------

By no  other way answer will come.

 

good dedicate your cash on this products. You can observe mobdro on pc In this article can be the program namely "MOBDRO" nice.

The ans would be 15 days.

For complete explanation on PERT and SD, you can read following articles

* PERT Formula and Use of Standard Deviation

* How to use PERT, CPM and Standard Deviation Together?