You have just used the network planning model and found the critical path length is 30 days and the variance of the critical path is 25 days. The probability that the project will be completed in 33 days or less is equal to (2 decimal accuracy)

A. 0.73 B. 0.55. C. 0.12 D. 0.27 E. 0.60.

The probability that the project will be completed in 33 days or less is equal to 0.73.

Step-by-step explanation:

The correct option is A:

Suppose that this is a normally distribution

The critical path length is 30 days.

i. e. Population mean = u=30

Variance: σ² = 25

Standard deviation = σ = √25 = 5

z-score: z = x-u / σ

The value of z corresponding to 33 = z = 33-30/5

z = 3/5

z = 0.6

Now, the p-value = P (x≤33) = P (z≤0.6) = 0.7257469

By rounding off we get,

0.73

Thus the probability that the project will be completed in 33 days or less is equal to 0.73.