A Building contractor is preparing a bid on a new construction project. Two othe

Question

A Building contractor is preparing a bid on a new construction project. Two other contractors will be submitting bids for the same project. Based on past bidding practices and the requirements of the projects, the bid from Contractor A can be described with a uniform distribution between $600,000 and $800,000, while the bid from Contractor B can be described with a normal distribution with a mean of $700,000 and standard deviation of $50,000.

1. If the building contractor submits a bud of $750,000, what is the probability that the building contractor will obtain the bid?

2. The building contractor is also considering bids of $765,000 and $775,000. If the building contract would like to bid such that the probability of winning the bid is about 0.80, what bid would you recommend? Repeat the simulation with bids of $765,000 and $775,000 to justify your recommendation.

Explanation / Answer

1) P(contractor A submitting a bid of $750,000 or lower = (750,000-600,000)/(800,000-600,000)= 0.75

2) P(contractor B submitting a bid of $750,000 or lower =

P(Z= (X -M)/sd) = P(Z= (750,000-70,000)/50000

P(Z=1) = 0.8413

P(both contractors A and B bidding less than $750,000

= 0.75x0.8413 = 0.631

So, the probability of the contractor getting the bid is 0.631

B) If considering a bid of $765,000

P(contractor A submitting less than $765,000)

= (765,000-600,000)/(200,000) = 0.825

P(contractor B submitting bid less than $765,000)

= P(Z=(765,000-700,000)/50000) = P(Z=1.3) = 0.9032

P(getting contract)= 0.825x0.9032 = 0.745

When bid of $775,000 is considered

P(contracter A submitiing bid less than $775,000) = 175/200

= 0.875

P(contractor B submitting bid less than $775,000)

= P(Z=75,000/50,000) = P(Z= 1.5) = 0.933

P(getting the bid) = 0.875x0.933 = 0.816

So, the contractor should go for a bid of $775,000 to have about 80% chance of winning the bid

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