Question

Jonah Arkfeld, a building contractor, is preparing a bid on a new construction project. Two other contractors will be submitting bids for the same project. Jonah has analyzed past bidding practices and the requirements of the project to determine the probability distributions of the two competing contractors. The bid from Contractor A can be described with a triangular distribution with a minimum value of $600,000, a maximum value of $800,000, and a most likely value of $725,000. The bid from Contractor B can be described with a normal distribution with a mean of $700,000 and a standard deviation of $50,000.

Required:
a. If Jonah submits a bid of $750,000, what is the probability that he will win the bid for the project?
b. What is the probability that Contractor A and Contractor B will win the bid, respectively?

Answer 1

Answer:

Step-by-step explanation:

From the given information:

Let assume that:

A be the distribution of contractor A and B be the distribution for contractor B

Then:

$B \sim N ( \mu,\sigma^2)$

$B \sim N ( 700,000, 50,000^2)$

A $\sim$ triangular distribution.

(a) Suppose that Jonah submits a bit of $750000;

The probability that Jonah win = P( B < 750000) * P(A < 750000)

$P(B < 750000) = P \Big ( \dfrac{B - 700000}{50000}< \dfrac{750000 - 700000}{50000} \Big )$

$P(B < 750000) = P \Big (Z < \dfrac{50000 }{50000} \Big )$

Then,

P(B < 750000) = P(Z < 1)

P(B < 750000) = 0.841

To simplify P(A < 750000); we take 800000 as 8, and 725000 as 7.25

$A \sim triangular \ distribution$

$f(a) = \left \{ {{\dfrac{2(a-6)}{2 \times 1.25} \ \ 6< a < 7.25 } \atop { \dfrac{2(8-a)}{2 \times 0.75 } \ \ 7.25 < a < 8}} \right.$

$P(A < 7.25) = \int \limits ^ {7.25}_{6} f(a) \ .da$

$= \int \limits ^{7.25}_{6} \ \dfrac{(a-6)}{1.25} \ da + \int \limits ^{7.5}_{7.25} \ \dfrac{(8-a)}{0.75} \ da$

$= \dfrac{1}{1.25} \Big [ \dfrac{a^2}{2}- 6a \Big ] ^{7.25}_{6} + \dfrac{1}{0.75} \Big [ 8a - \dfrac{a^2}{2} \Big ] ^{7.5}_{7.25}$

= 0.625 + 0.2083

= 0.833

∴

The probability that Jonah wins  =  P( B < 750000) * P(A < 750000)

= 0.841 × 0.833

= 0.70055

= 70.06%

The probability that Jonah will win a bid = 70.06%

P(A win the bid) = P (A > 750000 ) * P(B < 750000)

P(A win the bid) = 0.841 * (1- 0.833)

P(A win the bid) = 0.14044

P(A win the bid) = 14.04%

P(B win the bid) = P(A < 750000) * P( B > 750000)

P(B win the bid) = 0.833 * ( 1 - 0.841)

P(B win the bid) = 0.13245

P(B win the bid) = 13.25%

∴

"A" win the bid is 14.04% and "B" win the bid is 13.25% respectively.