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  1. mathematics
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  3. time​ (days) immediate time​ (d

Time​ (days) Immediate Time​ (days) Immediate Activity a m b ​Predecessor(s) Activity a m b ​Predecessor(s) A 55 55 77 long dash— H 44 44 66 ​E, F B 11 22 55 long dash— I 22 77 1010 ​G, H C 55 55 55 A J 22 44 77 I D 44 88 1313 A K 66 1010 1313 I E 11 1010 1717 ​B, C L 22 66 66 J F 11 55 77 D M 22 22 33 K G 22 66 99 D N 77 77 1212 ​L, M b) If the time to complete the...

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Question

Statistics

Time​ (days)

Immediate

Time​ (days)

Immediate

Activity

a

m

b

​Predecessor(s)

Activity

a

m

b

​Predecessor(s)

A

55

55

77

long dash—

H

44

44

66

​E, F

B

11

22

55

long dash—

I

22

77

1010

​G, H

C

55

55

55

A

J

22

44

77

I

D

44

88

1313

A

K

66

1010

1313

I

E

11

1010

1717

​B, C

L

22

66

66

J

F

11

55

77

D

M

22

22

33

K

G

22

66

99

D

N

77

77

1212

​L, M

b) If the time to complete the activities on the critical pathis normally​ distributed, then the probability that the criticalpath will be finished in 55 days or less​ =

Answer & Explanation Solved by verified expert
3.9 Ratings (602 Votes)

from abve:

as std deviaiton =(b-a)/6

hence

Activity Optimistic Likely Pessimistic Mean Std dev
A 5 5 7 5.33 0.33
B 1 2 5 2.33 0.67
C 5 5 5 5.00 0.00
D 4 8 13 8.17 1.50
E 1 10 17 9.67 2.67
F 1 5 7 4.67 1.00
G 2 6 9 5.83 1.17
H 4 4 6 4.33 0.33
I 2 7 10 6.67 1.33
J 2 4 7 4.17 0.83
K 6 10 13 9.83 1.17
L 2 6 6 5.33 0.67
M 2 2 3 2.17 0.17
N 7 7 12 7.83 0.83
Activity Early Start Early Finish Late Start Late Finish Slack
A 0.00 5.33 0.00 5.33 0.00
B 0.00 2.33 8.00 10.33 8.00
C 5.33 10.33 5.33 10.33 0.00
D 5.33 13.50 7.17 15.33 1.83
E 10.33 20.00 10.33 20.00 0.00
F 13.50 18.17 15.33 20.00 1.83
G 13.50 19.33 18.50 24.33 5.00
H 20.00 24.33 20.00 24.33 0.00
I 24.33 31.00 24.33 31.00 0.00
J 31.00 35.17 33.50 37.67 2.50
K 31.00 40.83 31.00 40.83 0.00
L 35.17 40.50 37.67 43.00 2.50
M 40.83 43.00 40.83 43.00 0.00
N 43.00 50.83 43.00 50.83 0.00

as acticvity on crtical path (with ) slack are =A ,C ,E ,H , I ,K M,N

thereofre project std deviation=sqrt(sum of variance of activiities on critical path)

=sqrt(11.194) =3.346

hence probability that the critical path will be finished in 55 days or less​ =P(X<55)

=P(Z<(55-50.83)/3.346)=P(Z<1.25)=0.8944
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