PERT probability: likelihood of meeting a 20-day target on the critical path A PERT network has three activities on the critical path with mean times 3, 8, and 6 days and standard deviations 1, 2, and 2 days respectively. What is the probability that the project will be completed within 20 days?

Introduction / Context:
PERT treats activity durations as random variables and often assumes the project duration on the critical path is approximately normal (by aggregation). Using the mean and variance of critical activities, we can estimate the probability of meeting a target completion time.

Given Data / Assumptions:

• Critical path activities: means = 3, 8, 6 days.
• Standard deviations = 1, 2, 2 days (assumed independent).
• Target completion time T = 20 days.

Concept / Approach:
Mean project duration on the critical path: mu = sum of means. Variance on the critical path: sigma^2 = sum of variances. Compute Z = (T - mu) / sigma and use the standard normal distribution to obtain the probability.

Step-by-Step Solution:
Compute mean: mu = 3 + 8 + 6 = 17 days.Compute variance: sigma^2 = 1^2 + 2^2 + 2^2 = 1 + 4 + 4 = 9.Compute standard deviation: sigma = sqrt(9) = 3 days.Compute Z-score: Z = (T - mu) / sigma = (20 - 17) / 3 = 1.0.Look up cumulative probability: Phi(1.0) ≈ 0.8413 → approximately 0.84.

Verification / Alternative check:
A normal table or calculator confirms that a Z of 1.0 corresponds to about 84% chance of finishing by the target.

Why Other Options Are Wrong:
0.50 implies T equals the mean (not true); 0.66 and 0.72 correspond to Z near 0.41–0.58; 0.95 corresponds to Z ≈ 1.64, which would require a later target.

Common Pitfalls:
Adding standard deviations instead of variances; forgetting to take the square root; using non-critical activities in the variance; rounding errors that move the probability substantially.

Final Answer:
0.84