In-Class Exercise: The Poisson Distribution

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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 autos 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?

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Solution Part a

• Let X be the number of surface flaws in a cars
  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.

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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.

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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).

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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).

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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 its true that P(Y 0) P(Z 0), it is
  not true that P(Y x) P(Z x) for x gt 0.
• So, we must use the Binomial distribution for
  Part c.