Respuesta :
Answer: 0.8186 Â
Explanation:
Given that;
activity To   Tm Tp   Te (V)^0.5       v
A Â Â Â Â 38 Â Â 50 62 Â Â 50 4 Â Â Â Â Â Â Â Â Â 16
B Â Â Â Â 90 Â Â 99 108 Â Â 99 3 Â Â Â Â Â Â Â Â Â Â 9
C Â Â Â Â 70 Â Â 80 90 Â Â 80 3.333333 Â Â 11.11111
D Â Â Â Â 19 Â Â 25 31 Â Â 25 2 Â Â Â Â Â Â Â Â Â Â 4
E Â Â Â Â 91 Â Â 100 115 Â Â 101 4 Â Â Â Â Â Â Â Â Â 16
F Â Â Â Â 62 Â Â 65 68 Â Â 65 1 Â Â Â Â Â Â Â Â Â Â 1
Expected duration Te = (4 × Tm + To + Tp ) / 6
Variance = ( Tp-To/6]²
variance of the critical path = 9+16 =25
SD of the critical path = ( var)^0.5 = 5
probability that the project will be completed within  210 days is given by
z = (210-200) / 5 = 2
which gives probability of 0.97725
Probability that the project will be completed within 195 days
z = (195-200) / 5 = -1
which corresponds to probability of 0.1586
Now required probability that project completes within 210 but before 195 days is given by
0.97725 - 0.1586 = 0.8186 Â