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Chapter 5: Problem 15

One hundred items are simultaneously put on a life test. Suppose the lifetimes of the individual items are independent exponential random variables with mean 200 hours. The test will end when there have been a total of 5 failures. If \(T\) is the time at which the test ends, find \(E[T]\) and \(\operatorname{Var}(T)\).

Short Answer

Expert verified
The expected test duration until 5 failures occur, \(E[T]\), is 10 hours, and the variance of the test duration, \(\operatorname{Var}(T)\), is 20 hours^2.

Step by step solution

01

\(\lambda_{item} = \dfrac{1}{200}\), [=[failure per hour]] #Step 2: Calculate the failure rate of the Poisson process# Since there are 100 items, we need to find the combined failure rate for the whole system. As the failures are independent the rates sum up. Thus, the failure rate for the entire set of 100 items is given by:

\(\lambda = 100 \cdot \lambda_{item} = \dfrac{100}{200} = \dfrac{1}{2}\), [=[failure per hour]] #Step 3: Calculate the expected value of the test duration# We need to find the expected time \(E[T]\) until the 5th failure occurs. This is the expected waiting time for 5 failures in a Poisson process. For a Poisson process, the expected waiting time for the \(k^{th}\) failure is given by \(\dfrac{k}{\lambda}\). In this case, we have:
02

\(E[T] = \dfrac{5}{\lambda} = \dfrac{5}{(1/2)} = 10\), [=[hours]] #Step 4: Calculate the variance of the test duration# The variance for waiting time until the \(k^{th}\) failure in a Poisson process is given by \(\dfrac{k}{\lambda^2}\). In our case, we can find the variance of \(T\) as following:

\(\operatorname{Var}(T)=\dfrac{5}{\lambda^2}=\dfrac{5}{(1/2)^2}=20\), [=[hours^(2)]] To summarize, the expected test duration until 5 failures occur is 10 hours, and the variance of the test duration is 20 hours^2.

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Most popular questions from this chapter

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