Time (days) Immediate Time (days) Immediate Activity a m b Predecessor(s) Activi
ID: 3069149 • Letter: T
Question
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 path is normally distributed, then the probability that the critical path will be finished in 55 days or less =
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
Explanation / Answer
from abve:
as std deviaiton =(b-a)/6
hence
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
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 0.83Related Questions
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