Can't see why Option B is correct. pls help

Submitted by tonsi on Wed, 11/28/2007 - 21:44.

The following three tasks form the entire critical path of the project network. The three estimates of each of these tasks are tabulated below. How long would the project take to complete expressed with an accuracy of one standard deviation?

Task  Optimistic  Most likely  Pessimistic  
A 15 25 47
B 12 22 35
C 16 27 32

A. 75.5
B. 75.5 +/- 5.08
C. 75.5 +/- 8.5
D. 75.5 +/- 2.83


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auto cleansed :)

Submitted by eugen on Mon, 01/07/2008 - 16:46.

auto cleansed 

(there is no button for delete) 

Can't see why Option B is correct. pls help

Submitted by anojv3 on Tue, 06/10/2008 - 13:43.
the option B is correct but the values is 75.5 + - 7.08 [(sqrt( sum of standard deviation of all the three tasks)) = 7.08] and not 5.08 Smile

probability

Submitted by vaalo on Fri, 06/27/2008 - 22:49.
the answer couldn't be 75.5+ -7.08 because the sum of the most likely estmates is 74, therefore it should be 74 + - 7.08

Probability calculation

Submitted by mlleeder on Wed, 08/06/2008 - 02:48.

75.5 looks right based upon the calculation for PERT:

(P + O + (4*M))/6

(114 + 43 + (4 * 74))/6 = 75.5

 But....

Regarding the post above with the standard deviation calculation:

I'm a bit confused by this - isn't the "standard deviation" calculation for this type of question/problem: (P-O)/6   ???

I have this listed in different study literature (one which I downloaded from this site).

Granted, it doesn't fit this answer, but where does it fit in?

 Should I be prepared to use the more traditional standard deviation formula?

 ..just wondering....Thanks!