how to find f(x) and how to deal with the questions?

Answered question

2021-11-16

Jones figures that the total number of thousands of miles that an auto can be driven before it would need to be junked is an exponential random variable with mean 20. Smith has a used car that he claims has been driven only 10,000 miles. If Jones purchases the car, what is the probability that she would get at least 20,000 additional miles out of it? Repeat under the assumption that the lifetime mileage of the car is not exponentially distributed, but rather is (in thousands of miles) uniformly distributed over (0, 40).

Answer & Explanation

Jeffrey Jordon

Jeffrey Jordon

Expert2021-11-18Added 2605 answers

Let A be exponential random variable that represents the number of thousands of miles that a used auto can be
driven, Xexp(120)
So what we want to calculate is probability that the car will cross 30 thousand miles if we have that it has already crossed 10 thousand miles:

P(X<30|X>10)=P(X>20+10|X>10)=P(X>20)=e12020=0.368

Now let X be uniformly distriduted, XU(0,40). Now we have conditional probability:

P(X>30|X>10)=P(X>30)P(X>10)=1P(X30)1P(X10)=130/40110/40=13

user_27qwe

user_27qwe

Skilled2021-12-03Added 375 answers

Let X be the number of miles the car can be driven before being junked.

(a) First, let XExp(12). We know that the car has been driven 10,000 miles and we want to find the proba- bility that it will last an additional 20,000 miles. We use the memoryless property of exponential distribution, where t=10000 and s=20000.

Then, P(X10000+20000|X>10000)=P(X20000)

P(X30000|X>10000)=P(x20000)

=20e120xdx

=e1

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