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​ =

Answer 1

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