Question

a) the expected (estimated) time for activity c is 13.67 weeks (round your response to two decimal places.)
b) the variance for activity c is 2.78 weeks (round your response to two decimal places.)
c) based on the calculation of the estimated times, the critical path is a - c - f - h - j - k
d) the estimated time for the critical path is 45.34 weeks. (round your response to two decimal places.)
e) the activity variance along the critical path is 12.06 weeks. (round your response to two decimal places.)
f) using the standard normal table, the probability that hopkins hospitality will finish the project in 36 weeks or less is (enter as a probability and round your response to three decimal places.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x - \mu}{\sigma}\), where \(x = 36\) (the target time), \(\mu=45.34\) (the estimated time of the critical path), and \(\sigma=\sqrt{12.06}\approx3.47\) (the standard deviation of the critical path, since variance \(V = 12.06\) and \(\sigma=\sqrt{V}\)).

$$z=\frac{36 - 45.34}{3.47}=\frac{- 9.34}{3.47}\approx - 2.7$$

Step2: Find the probability from the standard - normal table

Looking up \(z=-2.7\) in the standard - normal table. The standard - normal table gives the cumulative probability \(P(Z\leq z)\). For \(z = - 2.7\), the cumulative probability is \(0.0035\)

Answer:

\(0.004\)