Question

A project manager has compiled a list of major activities that will be required to install a computer information system in her firm. The list includes estimated completion times for activities and precedence relationships.
Use standard deviation table.

Activity	Immediate Predecessor	Estimated Times (weeks)
A	—	2-4-6
D	A	6-8-10
E	D	7-9-12
H	E	2-3-5
F	A	3-4-8
G	F	5-7-9
B	—	2-2-3
I	B	2-3-6
J	I	3-4-5
K	J	4-5-8
C	—	5-8-12
M	C	1-1-1
N	M	6-7-11
O	N	8-9-13
End	H, G, K, O

  
If the project is finished within 26 weeks of its start, the project manager will receive a bonus of $1,000; and if the project is finished within 27 weeks of its start, the bonus will be $500. Find the probability of each bonus. (Round Mean, Standard Deviation, z-value to 2 decimal places and Probability to 4 decimal places.)

Path	Mean	Std. Dev.
a-d-e-h
a-f-g
b-i-j-k
c-m-n-o

  
Probability ($1,000)   

Probability ($500)            

Answer 1

We find the expected time and Standard Deviation for each activity as shown below;

Path A-D-E-H:

Expected duration = 4 + 8 + 9.17 + 3.17 = 24.34

Mean Standard Deviation = = = 1.35

Path A-F-G:

Expected duration = 4 + 4.50 + 7 = 15.5

Mean Standard Deviation = = = 1.26

Path B-I-J-K:

Expected duration = 2.17 + 3.33 + 4 + 5.33 = 14.83

Mean Standard Deviation = = = 1.01

Path C-M-N-O:

Expected duration = 8.17 + 1.00 + 7.50 + 9.50 = 26.17

Mean Standard Deviation = = = 1.66

As seen from above, path C-M-N-O are critical. Duration = 26.17 weeks Standard Deviation = 1.66

Hence, we find Z value in comparison to this path as :

For 26 weeks:

Z = = = (-0.102)

For Z = (-0.102), Probability = 0.45220= 45.22%

Probability ($1,000) = 45.22%

For 27 weeks:

Z = = = 0.5

For Z = 0.5, Probability = 0.6915 = 69.15%

Probability ($5,000) = 69.15%

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

In case of any doubt, please ask through the comment section before Upvote/downvote.