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 $10,500 will be accepted. Assume that the competitor's bid x is a random variable that is uniformly distributed between $10,500 and $14,500.

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

Suppose you bid $14,000. What is the probability that your bid will be accepted (to 2 decimals)?

What amount should you bid to maximize the probability that you get the property (in dollars)?

What is the expected profit for this bid (in dollars)?

Answer 1

for uniform distribution parameter:a =10500and b=14500

a)probability that 12000 bid be accepted =P(X<12000)=(x-a)/(b-a)=(12000-10500)/(14500-10500)=

0.38

b)probability that 14000 bid be accepted =P(X<14000)=(x-a)/(b-a)=(14000-10500)/(14500-10500)=

0.88

c)amount to  maximize the probability that you get the property =14500

( pleas provide hpw much money detail to get the correct answer of expected profit)