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, bids from the other contractors can be described by the following probability distributions:

If required, round your answers to three decimal places.

If the building contractor submits a bid of $710,000, what is the probability that the building contractor will obtain the bid? Use an Excel worksheet to simulate 1,000 trials of the contract bidding process.

The probability of winning the bid of $710,000 =

The building contractor is also considering bids of 735,000 and $745,000. If the building contractor would like to bid such that the probability of winning the bid is about 0.7, what bid would you recommend? Repeat the simulation process with bids of $735,000 and $745,000 to justify your recommendation.

The probability of winning the bid of $735,000 =  
The probability of winning the bid of $745,000 =  

The reccomendation would be to choose the bid of $

Contractor Probability Distribution of Bid A Uniform probability distribution between $570,000 and $770,000 B Normal probability distribution with a mean bid of $670,000 and a standard deviation of $47,000

Explanation / Answer

You can use excel as well. But to visualize the code, I have used R here.

CODE:

a=round(runif(1000,570000,770000),3)
b=round(rnorm(1000,670000,47000),3)
count=0
for(i in 1:1000) {
if(a[i] < 710000 && b[i] < 710000) count =count+1
}
count
py = count/1000
py

count=0
for(i in 1:1000) {
if(a[i] < 735000 && b[i] < 735000) count =count+1
}
count
p1 = count/1000
p1

count=0
for(i in 1:1000) {
if(a[i] < 745000 && b[i] < 745000) count =count+1
}
count
p2 = count/1000
p2

OUTPUT:

> a=round(runif(1000,570000,770000),3)
> b=round(rnorm(1000,670000,47000),3)
> count=0
> for(i in 1:1000) {
+ if(a[i] < 710000 && b[i] < 710000) count =count+1
+ }
> count
[1] 566
> py = count/1000
> py
[1] 0.566
>
> count=0
> for(i in 1:1000) {
+ if(a[i] < 735000 && b[i] < 735000) count =count+1
+ }
> count
[1] 758
> p1 = count/1000
> p1
[1] 0.758
>
> count=0
> for(i in 1:1000) {
+ if(a[i] < 745000 && b[i] < 745000) count =count+1
+ }
> count
[1] 829
> p2 = count/1000
> p2
[1] 0.829

therefore,

Prob[winning bid 710,000] = .566

Prob[winning bid 735,000] = .758

Prob[winning bid 745,000] = .829

As the contractor want to bid such that prob is about .7 then 735,000 is a recommended bid since the bid value is lesser than that of 745,000 and provides a winning prob nearly .7

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