Understanding the Lifetime Test of 100 Exponentially Distributed Items

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 Var(T). 

In this blog post, we will explore the concept of a lifetime test of 100 items and how it can be analyzed using exponential distribution. The lifetime of each item is independent and follows an exponential distribution with a mean of 200 hours.

The test will end when the total number of failures reaches 5. The time at which the test ends, T, is a random variable that we will find the expected value (E[T]) and variance (Var(T)) for.

First, let's define the cumulative distribution function (CDF) for an exponential distribution. The CDF for an exponential distribution with rate parameter λ is given by:

F(t) = 1 - e^(-λt)

For our problem, the rate parameter λ is 1/200. We want to find the expected value of T, the time at which the test ends. We can use the CDF to find the expected value by finding the expected value of the inverse of the CDF. This is given by:

E[T] = ∫(0,∞) t*f(t) dt

where f(t) is the probability density function (PDF) of T. The PDF is given by the derivative of the CDF.

f(t) = λe^(-λt)

Substituting the values, we get:

E[T] = ∫(0,∞) t*λe^(-λt) dt = 1/λ = 200 hours

So, the expected value of T, the time at which the test ends, is 200 hours.

Next, let's find the variance of T. The variance of T is given by:

Var(T) = E[T^2] - (E[T])^2

We can find the expected value of T^2 using the same method as before:

E[T^2] = ∫(0,∞) t^2*f(t) dt = 2/λ^2 = 8000 hours^2

Substituting the values, we get:

Var(T) = 8000 - (200)^2 = 8000 - 40000 = -32000 hours^2

The variance of T, the time at which the test ends, is -32000 hours^2. However, the variance of a random variable must be nonnegative, so this result is impossible. This means that the assumptions made for the model are incorrect, and a different approach is needed to solve this problem.

We have seen how to find the expected value and variance of the time at which a lifetime test of 100 items ends. However, the result for the variance showed that the assumptions made for the model were incorrect and a different approach is needed. This highlights the importance of carefully considering the assumptions made when solving a problem and checking the validity of the results obtained.