The new director of special events at a large university has decided to complete
ID: 371531 • Letter: T
Question
The new director of special events at a large university has decided to completely revamp graduation ceremonies. Toward that end, a PERT chart of the major activities has been developed. The chart has five paths with expected completion times and variances as shown in the table. Graduation day is 16 weeks from now.
Use Table B and Table B1.
Assuming the project begins now, what is the probability that the project will be completed before: (Round your z-value to 2 decimal places and all intermediate probabilities to 4 decimal places. Round your final answers to 4 decimal places.)
a. Graduation time?
Probability for graduation time
b. The end of week 15?
Probability at the end of week 15
c. The end of week 13?
Probability at the end of week 13
(weeks) Variance A 10 1.21 B 8 2.00 C 12 1.00 D 15 2.89 E 14 1.44
Explanation / Answer
Path
Expected Duration
Variance
(weeks)
A
10
1.21
B
8
2.00
C
12
1.00
D
15
2.89
E
14
1.44
Steps
There are five paths (not activities). The path which takes the longest time is the critical path. Here it is path D with 15 weeks. This is the mean completion time of the project.
Mean completion time of the project =15 weeks
Standard deviation of the project= (variance ) =(2.89 ) = 1.7
Find z ( the standardized deviation of the project from mean) and find the corresponding probability(area under curve) for Z using table
a.
Probability that the project will be completed before graduation-16 weeks
Z=Specified time Path mean /Path standard deviation =16-15/1.7 =0.5882
From z table (standard normal distribution table , we find the probability for the Z value as 0.5882
0.719 or 71.9%
B.
Probability that the project will be completed before 15 weeks
Z=Specified time Path mean /Path standard deviation =15-15/1.7 =0.5000
From z table (standard normal distribution table , we find the probability for the Z value as 0.5000
50%
C.
Probability that the project will be completed before 13 weeks
z =Specified time Path mean /Path standard deviation =13-15/1.7 =-1.1764
From z table (standard normal distribution table, we find the probability for the Z value as 0.1210
12.1%
Path
Expected Duration
Variance
(weeks)
A
10
1.21
B
8
2.00
C
12
1.00
D
15
2.89
E
14
1.44
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