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and is equal to the square of the population's standard deviation. The variances in which we are interested are the variances of the activities on the critical path.

The critical path of our example includes activities a, d, and j. From Table 8-2 we find that the variances of these activities are 4, 25, and 4, respectively; and the variance for the critical path is the sum of these numbers, 33 days. Assume, as above, that the PM has promised to complete the project in 50 days. What are the chances of meeting that deadline? We find the answer by calculating Z, where

D = the desired project completion time fi = the critical time of the project, the sum of the TEs for activities on the critical path o1^2 = the variance of the critical path, the sum of the variances of activities on the critical path

Z = the number of standard deviations of a normal distribution (the standard normal deviate)

Z, as calculated above, can be used to find the probability of completing the project on time.

Using the numbers in our example, D = 50, |x = 43, and a^2 = 33 (the square root of 2 is 5.745), we have

Table 8-4 Cumulative (Single Tïtil) Probabilities of the Normal Probability

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