Question

6. A new integrated computer system is to be installed worldwide for a major corporation. Bids on this project are being solicited, and the contract will be awarded to one of the bidders. As a part of the proposal for this project, bidders must specify how long the project will take. There will be a significant penalty for finishing late. One potential contractor determines that the average time to complete a project of this type is 42 weeks with a standard deviation of 6 weeks. The time required to complete this project is assumed to be normally distributed.

1. If the due date of this project is set at 42 weeks, what is the probability that the contractor will have to pay a penalty (i.e., the project will not be finished on schedule)?
2. If the due date of this project is set at 43 weeks, what is the probability that the contractor will have to pay a penalty (i.e., the project will not be finished on schedule)?
3. If the bidder wishes to set the due date in the proposal so that there is only a 5% chance of being late (and consequently only a 5% chance of having to pay a penalty), what due date should be set?

Answer 1

Let X be the time taken in weeks to complete a project, then X ~ N(42,6²). We will use the result Z= and the z-values are obtained from standard normal probability tables and the contractor will have to pay a penalty if the time taken to complete the project is greater than the due date.

(a) If the due date of this project is set at 42 weeks, the probability that the contractor will have to pay a penalty is given by

P (X>42) = P (Z > (42-42)/6) = P (Z>0) = 0.5

(b) If the due date of this project is set at 43 weeks, the probability that the contractor will have to pay a penalty is given by

P(X>43) = P (Z > (43-42)/6 ) = P(Z> 0.16) = 1 - P (Z< 0.16) = 1-0.5635 = 0.4365

(c) Let a be the due date in weeks to be selected so that there is only a 5% chance of being late.

Hence we have to calculate the value of a such that P (X>a) =0.05 or P (Z > (a-42)/6) = 0.05

Thus (a-42)/6 = 1.6449 => a = 1.6449*6 +42 = 51.87 or a 52 ( P (Z>z) = 0.05 => z=1.6449 from standard normal probability tables)

Hence the bidder should set the due date as 52 weeks so that there is only a 5% chance of being late.