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 $10,100 will be accepted. Assume that the competitor's bid x is a random variable that is uniformly distributed between $10,100 and $14,700. Suppose you bid $12,000. What is the probability that your bid will be accepted (to 2 decimals)

Answer 1

`P(X<12000)`

And for this case we can use the cumulative distribution function given by:

`P(X\leq x) =(x-a)/(b-a), a \leq x \leq b`

And using this formula we have this:

`P(X<12000)= (12000-10100)/(14700-10100)= 0.41`

Then we can conclude that the probability that your bid will be accepted would be 0.41

Explanation:

Let X the random variable of interest "the bid offered" and we know that the distribution for this random variable is given by:

`X \sim Unif( a= 10100, b =14700)`

If your offer is accepted is because your bid is higher than the others. And we want to find the following probability:

`P(X<12000)`

And for this case we can use the cumulative distribution function given by:

`P(X\leq x) =(x-a)/(b-a), a \leq x \leq b`

And using this formula we have this:

`P(X<12000)= (12000-10100)/(14700-10100)= 0.41`

Then we can conclude that the probability that your bid will be accepted would be 0.41