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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)

Respuesta :

Answer:

[tex] P(X<12000)[/tex]

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

[tex] P(X\leq x) =\frac{x-a}{b-a}, a \leq x \leq b[/tex]

And using this formula we have this:

[tex] P(X<12000)= \frac{12000-10100}{14700-10100}= 0.41[/tex]

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

Step-by-step explanation:

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

[tex] X \sim Unif( a= 10100, b =14700)[/tex]

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

[tex] P(X<12000)[/tex]

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

[tex] P(X\leq x) =\frac{x-a}{b-a}, a \leq x \leq b[/tex]

And using this formula we have this:

[tex] P(X<12000)= \frac{12000-10100}{14700-10100}= 0.41[/tex]

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

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