Exponential Distribution Problem

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RyNye

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Sep 22, 2013
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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.
Click to expand...

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?
 
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.
 
HallsofIvy said:
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.
Click to expand...

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?
 
RyNye said:
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?
Click to expand...
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)
 
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