The Expected Value

The Problem

Let X be a random variable following a Poisson distribution. All Poisson distributions have just one parameter: average rate, λ (lambda). For this problem, let X have rate parameter λ = 15.67. An example of where such a distribution may arise is the following:

Statistics, Inc., makes candy. Unfortunately, they are not too good at it. Every week, an average of 15.67 pieces are defective. These defective pieces will not kill a person, but they will cause the person’s left index finger to turn scarlet. Let X be the number of defective pieces of candy manufactured in a specified week.

From the description, we can tell that X has a Poisson distribution with rate parameter λ = 15.67. We know X has this distribution because the random variable is the number of successes measured over time (or space).

For those who like pictures, here is a graphic of the probability mass function of X. Note that it is skewed positive (right), that its sample space is non-negative, and that there is no upper bound to its sample space. In other words, S = {0, 1, 2, …}.

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Continuing the candy example, let us calculate the expected number of defective pieces of candy produced each week. That is, calculate E[X] (or μ).

Your Answer

Another Poisson Distribution

Would you like to continue working on this topic? If so, click here for another Poisson distribution.

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