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Answers only Please don’t make it confusing Please be clear Please write or type neatly A bulding contractor is preparing ? bid on a new construction project. Two other contractors will be submitting bids for the same project Based on past bidding practices, bids from the other contractors can be described by the following probability distributions Probability Distribution of Bid Uniform probability distribution between $590,000 and $790,000 Normal probability distribution with a mean bid of $690,000 and a standard deviation of $49,000 If required, round your answers to three decimal places. a. If the building contractor submits a bid of $730,000, what is the probability that the building contractor will abtain the bid? Use an Excel worksheet to simulate 1,000 trials of the contract bidding process. The probability of winning the bid of $730,000-0.sss x b. The building contractor is also considering bids of 755,000 and $765,000. If the building contractor would like to bid such that the probability of winning the bid is about 0.6, what bid would you recommend? Repeat the simulation process with bids of $755,000 and $765,000 to justify your recommendation The probability of winning the bid of $755,000-| 0.749 x The probabillity of winning the bid of $765,0D0-0.820 X The reccomendation would be to choose the bid of $ 755000
Explanation / Answer
As Contractor A bid follows a uniform distribution, the probability that the bid is more than contractor A is:
P(A) = (730000 - 590000)/(790000 - 590000)
P(A) = 0.7
As Contractor B bid follows a normal distribution, the probability that the bid is more than contractor B is:
P(z<=x) where z = (730000 - 690000)/49000
z = 0.816
Hence, P(z<=x) = P(B) = 0.792
Therefore, probability to win the bid from both contractors = P(A). P(B) = 0.7*0.792 = 0.555
B) Doing the same procedure for the bid of 755000,
P(A) = 0.825
P(B) = 0.907
P(win) = 0.749
The probability of winning the bid of $765000 = P(A). P(B) = 0.875 * 0.937 = 0.820
The recommendation would be to choose the bid of $755000.
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