Question

A PERT network consists of 60 activities, and 20 of these are critical activities. The length of the critical path is 120 days. The sum of all activity variances is 64; while the sum of the variances along the critical path is 36.

The probability that this project can be completed within 108 and 120 days is:

1. More than 0.4675 but less than or equal to 0.4695
2. More than 0.4695 but less than or equal to 0.4700
3. More than 0.4700 but less than or equal to 0.4775
4. More than 0.4775 but less than or equal to 0.4780
5. None of the above.

Answer 1

Solution:
Given in the question
Length of critical path or Expected time of project completion ($\mu$) = 120
Variance of Critical path = 36
Standard deviation ($\sigma$) = sqrt(Variance) = sqrt(36) = 6
We need to calculate probability that this project can be completed within 108 and 120 days, which can be calculated as
P(108<X<120) = P(X<120) - P(X<108)
Here we will use standard normal distribution, First we will calculate Z-score which can be calculated as
Z-score = (X-$\mu$)/$\sigma$ = (120-120)/6 = 0
Z = (108-120)/6 = -2
From Z table we found p-value
P(108<X<120) = P(X<120) - P(X<108) = 0.5 - 0.0228 = 0.4772
So there is 47.72% probability that this project can be completed within 108 and 120 days.