Question

Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that the highest bid in excess of $9,500 will be accepted. Assume that the competitor's bid x is a random variable that is uniformly distributed between $9,500 and $14,600.

Answer 1

a. Suppose you bid $11,500. What is the probability that your bid will be accepted?

b. Suppose you bid $13,500. What is the probability that your bid will be accepted?

Answer:

a. 0.392

b. 0.784

Step-by-step explanation:

Given

a = 9,500 , b = 14,600

The probability density function is given by 1 divided by the interval between a and b.

f(x) = 1/(b - a)

f(x) = 1/(14,600 - 9,500)

f(x) = 1/5100

f(x) = 0.000196

a. Suppose you bid $11,500. What is the probability that your bid will be accepted?

This is given by the integration of f(x) over the interval in the probability

I.e.

P(x < 11,500) = Integral of 0.000196dx, where upper bound = 11,500 and lower bound = 9,500

Integrating 0.000196dx gives

0.000196x introducing the upper and lower bound.

We get

0.000196(11,500 - 9,500)

= 0.392

b. Suppose you bid $13,500. What is the probability that your bid will be accepted?

This is given by the integration of f(x) over the interval in the probability

I.e.

P(x < 13,500) = Integral of 0.000196dx, where upper bound = 13,500 and lower bound = 9,500

Integrating 0.000196dx gives

0.000196x introducing the upper and lower bound.

We get

0.000196(13,500 - 9,500)

= 0.784