The following problem is taken from the homework problems in Probability and Statistics by Mendenhall, Beaver and Beaver. As you can see it deals with the very popular and useful Poisson distribution.
The key for effectively using the Poisson Distribution is to recognize or be able to calculate from givens the value of lambda. Lambda behaves as the mean or average number of occurrences of some event over time or in some area.
The easiest way to solve this particular problem would work out the complement which would be 1 minus the probability of 0,1,2,3,4,or 5 bacteria being present in the specimen. The work is shown below via the Maple CAS where the Poisson probability distribution was defined in terms lambda and x. In this case x represents the number of bacteria present in particular specimen.
For this particular problem the value lambda turns out to be 2 which is the mean count of some type of bacteria living in a water specimen. The question is what is the probability that a water supply with this particular value of lambda would have more than 5 bacteria present in a particular specimen? The probability of more than 5 bacteria occurring in an arbitrary specimen is approximately 0.017.
This means in only 17 specimens out of 1000 would a safety violation occur that would jeopardize the health of a town with this particular water supply. Put another way the concentration of this particular type of harmful bacteria is almost always very low!
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If a drop of water is placed on a slide and examined under a microscope, the number x of a particular type of bacteria present has been found to have a Poisson probability distribution. Suppose the maximum permissible count per water specimen for this type of bacteria is 5. If the mean count for your water supply is two and you test a single specimen, is it likely that the count will exceed the maximum permissible count? Explain