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 $9,600 will be accepted. Assume that the competitor's bid x is a random variable that is uniformly distributed between $9,600 and $14,500. Suppose you bid $12,000. What is the probability that your bid will be accepted (to 2 decimals)?

Answer 1

Answer:

0.49 = 49% probability that your bid will be accepted

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X lower than x is given by the following formula.

$P(X \leq x) = \frac{x - a}{b-a}$

Uniformly distributed between $9,600 and $14,500.

This means that $a = 9600, b = 14500$

Suppose you bid $12,000. What is the probability that your bid will be accepted (to 2 decimals)?

Your bid will be accepted if the competitor's bid X is lower than 12000. So

$P(X \leq 12000) = \frac{12000 - 9600}{14500 - 9600} = 0.49$

0.49 = 49% probability that your bid will be accepted