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In-Class Exercise: The Poisson Distribution

In-Class Exercise: The Poisson Distribution. 3-105. The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a man of 0.05 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.

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In-Class Exercise: The Poisson Distribution

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  1. In-Class Exercise: The Poisson Distribution • 3-105. The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a man of 0.05 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel. • What is the probability that there are no surface flaws in an auto’s interior? • If 10 cars are sold to a rental car company, what is the probability that none of the 10 cars has any surface flaws? • If 10 cars are sold to a rental car company, what is the probability that at most one car has any surface flaws?

  2. Solution: Part a • Let X be the number of surface flaws in a car’s interior. • Since there are an average of 0.05 flaws per square foot and a car interior contains of 10 square feet of the material, X is Poisson random variable with a mean of  = 0.5.

  3. Solution: Part b • Let Y be the number of surface flaws in a fleet of 10 cars. • Since there are an average of 0.05 flaws per square foot and one car interior contains of 10 square feet of the material, Y is Poisson random variable with a mean of  = 5.

  4. Solution: Part b • Let Z be the number of cars that have surface flaws in a fleet of 10 cars. • In Part a we found that the probability that a car has no surface flaws is e-0.5. • Therefore, the probability that a car has any surface flaws is (1-e-0.5) • If we treat the fleet of cars as sequence of 10 Bernoulli trials, then Z is a binomial random variable with n = 10 and p = (1-e-0.5).

  5. Solution: Part c • Let Z be the number of cars that have surface flaws in a fleet of 10 cars. • In part a we found that the probability that a car has no surface flaws is e-0.5. • Therefore, the probability that a car has any surface flaws is (1-e-0.5) • If we treat the fleet of cars as sequence of 10 Bernoulli trials, then Z is a binomial random variable with n = 10 and p = (1-e-0.5).

  6. Solution: Part c • Notice that the random variable Y counts the number of surface flaws in 100 square feet of plastic panel. • This is equivalent to 10 car interiors, however the event that there are say 6 surface flaws in 100 square feet of plastic panel can be realized in many different ways. For example • All 6 flaws are in the first 10 square feet (i.e., they are all in the first car) • 3 flaws are in the first 10 square feet, 2 flaws are in the second 10 square feet, and 1 is in the third 10 square feet (i.e., there are flaws in the first 3 cars). • While it’s true that P(Y = 0) = P(Z = 0), it is not true that P(Y = x) = P(Z = x) for x > 0. • So, we must use the Binomial distribution for Part c.

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