Question

A gentleman farmer maniacally works on academic projects to raise funds for his true passion, farming. He has identified seven tasks and developed three-time estimates (all in days) of the next big project on the farm, replacing the fence at the comedy pasture. Tasks, precedence requirements, and time estimates are shown in the table. What is the probability that the project is completed in 31 days?

Activity	Optimistic time	Most Likely time	Pessimistic time	Predecessor
Corner Posts	4	5	6	-
String Line	7	8	9	-
Dig Holes	10	12	14	Corner Posts
Plant Posts	11	14	17	String Line
Weld Top Bar	8	11	14	String Line
Paint	9	11	13	Dig Holes, Plants Posts
Tack Wire	9	12	16	Weld Top Bar

Answer 1

Answer: the probability that the project is completed in 31 days= 0.0544

Explanation:

Activity	Optimistic time-a	Expected completion time-m	Pessimistic time-b	Expected time= (a+4*m+ b)/6	Variance, (sigma)^2= (b-a/6)^2
A	4	5	6	5.00	0.11
B	7	8	9	8.00	0.11
C	10	12	14	12.00	0.44
D	11	14	17	14.00	1.00
E	8	11	14	11.00	1.00
F	9	11	13	11.00	0.44
G	9	12	16	12.17	1.36

For critical path

Paths	Duration
ACF	28.00
BDF	33.00
BEG	31.17

BDF is the longest path- critical path, with duration 33 days

For probability

step 1	we will find the variance of the tasks which lie on critical path				1.56
step 2	mean project time (u) of critical path is=				33.00
step 3	Required completion time is				31
step 4	standard deviation= sqrt(variance)=				1.247
step 5	because Z= (given completion time- u)/standard deviation				-1.604
step 6	P(z)= using NORM.S.DIST(z,true)				0.0544

The probability is 0.0544

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