Exponential Distribution Problem

[RyNye]

Hi everyone,

So in a practice exam I have been faced with the following problem:

An old machine is installed in a factory. The time until the machine breaks down follows an exponential distribution with a mean of 3 hours. After a breakdown the machine is repaired and the time to the next break down continues to follow an exponential distribution with a mean of 3 hours.
a) What is the probability the machine breaks down in less than 4 hours?
b) Suppose the machine breaks down and is repaired at 10:00am. Then the repairman takes a break until noon and is unavailable to repair the machine again if it breaks down again during that period. If the machine has a breakdown from 10:00am-noon the repairman will repair the machine at noon. If the machine has a breakdown at a time T after noon the repairman will repair the machine at time T. Let X be the random variable that depicts the time when the machine is repaired after 10:00am. Find the expected value of X.

Part a is pretty easy, I think. This is just an exponential distribution with a lambda value of 1/3. (The mean of the distribution is 3 hours, so that is the expected value, and E(X) = 1/lambda).

So we get P(X<4) = 1 - e^(4*1/3) ~= 0.74
Is that correct? I'd show my work but I don't know how to type the integral and such on the forum.

It is part b that I am stuck on. The repairman is going to repair the machine either at noon or sometime after. So X (if x is time from 10am) has to be greater than or equal to 2 hours. Time T after noon follows the original exponential distribution - due to the loss of memory property, the probability doesn't change just because we know that has to be greater than 2. But I don't understand what to do next. Isn't the expected value just 3? That's the mean of the original exponential distribution, and it shouldn't change, right?

 

[HallsofIvy]

1. Find the probability the machine breaks down before noon. In that case the machine will be repaired at noon so X= 2 (hours after 10:00). Find the probability times 2.

2. Find the probability the machine breaks down T hours after 10:00 with T> 2. Find the integral of probability time T for T> 2.

3. Add those two values.

 

[RyNye]

1. Find the probability the machine breaks down before noon. In that case the machine will be repaired at noon so X= 2 (hours after 10:00). Find the probability times 2.

2. Find the probability the machine breaks down T hours after 10:00 with T> 2. Find the integral of probability time T for T> 2.

3. Add those two values.

I don't see how that is right. First of all, T is time past noon, X is time past 10. So X has to be greater than or equal to 2. T is just going to be greater than 0.

If X is the random variable for time past 10am, the probability for it being less than 2 has to be 0. So this would just be the integral of the exponential function from 2 to infinity, right?

 

[DrPhil]

Hi everyone,

So in a practice exam I have been faced with the following problem:



Part a is pretty easy, I think. This is just an exponential distribution with a lambda value of 1/3. (The mean of the distribution is 3 hours, so that is the expected value, and E(X) = 1/lambda).

So we get P(X<4) = 1 - e^(4*1/3) ~= 0.74
Is that correct? I'd show my work but I don't know how to type the integral and such on the forum.

It is part b that I am stuck on. The repairman is going to repair the machine either at noon or sometime after. So X (if x is time from 10am) has to be greater than or equal to 2 hours. Time T after noon follows the original exponential distribution - due to the loss of memory property, the probability doesn't change just because we know that has to be greater than 2. But I don't understand what to do next. Isn't the expected value just 3? That's the mean of the original exponential distribution, and it shouldn't change, right?

I will try to rephrase what HallsofIvy said.

1) Find P(X<2). In this case the time from 10am till repair is X=2 hours

2) Otherwise, clock starts over at noon, with expected failure at 3pm, that is, X=5 hours

3) E[X] = P(X<2)*(2 h) + [1 - P(X<2)]*(5 h)