Expected Time = (a + 4m + b)/6
Variance = (b – a)^2/36
The z-value for the normal distribution a T =25 is given by Z = (25-22)/1.697 =1.77
Variance (\sigma^2) of the critical path activities: 1.78+0.11+ 0.44+0.11+0.11+0.11+0.11 =2.78
Standard Deviation = \sqrt{2.78} = 1.67
Using Z score equation
Using Z=1.77 and the standard normal distribution table
We find that the probability of the project meeting the 25 week deadline is (PZ <= 1.77) = 0.9616 = 96.16%
1.697weeks Z = (25-22)/1.697 =1.77