Other Scheduling Methods - Construction Project Scheduling and Control - page 301

day 30 or earlier, or the same as the probability that the project will
finish by the end of day 30:

Z =( 30 − 32 )∕ 2 . 972 =− 0 . 67

and

Pr ( T s ≤ 30 )= 1 − 0 . 749 = 0 . 251 = 25 . 1 %

SeeFigure11.7.

Figure 11.7 Solution to example 11.1, part 6

7. The probability that the project will finish at least 2 days late . This prob-
ability is the same as the probability that the project will finish by day
34 or later, or the same as the probability that the project will finish
after day 33:

Pr ( T s > 33 )= 1 − Pr ( T s ≤ 33 )

For P r ( T S ≤ 33 )

Z =( 33 − 32 )∕ 2 . 972 = 0 . 34

and

P r ( T s ≤ 33 )= 0 . 633 = 63 . 3 %

Thus,

Pr ( T s > 33 )= 1 − 0 . 633 = 0 . 367 = 36 . 7 %