Question

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Question 25 The number of surface flaws in a plastic roll used in the interior of automobiles has a Poisson distribution with a mean of 0.06 flaw per square foot of plastic roll. Assume an automobile interior contains 9 square feet of plastic roll. Round your answers to four decimal places (e.g. 98.7654) (a) What is the probability that there are no surface flaws in an auto's interior? (b) If 13 cars are sold to a rental company, what is the probability that none of the 13 cars has any surface flaws? (c) If 13 cars are sold to a rental company, what is the probability that at most 3 cars has any surface flaws?

Answer 1

(a)

Let X be the number of surface flaws in a 9 square feet plastic roll. Then X ~ Poisson($\lambda$ = 9 * 0.06 = 0.54)

Probability that there are no surface flaws in an auto's interior = P(X = 0)

= exp(-0.54) * 0.54⁰ / 0! = exp(-0.54) = 0.5827

(b)

From part (a), Probability that there are no surface flaws in a car = 0.5827

Let Y be the number of cars that has no flaws. Then Y ~ Binomial( n = 13, p = 0.5827)

Probability that none of the cars has flaws = P(Y = 13)

= ¹³C₁₃ * 0.5827¹³ * (1 - 0.5827)^13-13

= 0.5827¹³

= 0.0009

(c)

From part (a), Probability that there are any surface flaws in a car = 1 - 0.5827 = 0.4173

Let Z be the number of cars that has flaws. Then Z ~ Binomial( n = 13, p = 0.4173)

Probability that at most 3 of the cars has flaws = P(Z $\le$ 3) = P(Z = 0) + P(Z = 1) + P(Z = 2) + P(Z = 3)

= ¹³C₀ * 0.4173⁰ * (1 - 0.4173)^13-0 + ¹³C₁ * 0.4173¹ * (1 - 0.4173)^13-1 + ¹³C₂ * 0.4173² * (1 - 0.4173)^13-2 + ¹³C₃ * 0.4173³ * (1 - 0.4173)^13-3

= 0.5827¹³ + 13 * 0.4173 * 0.5827¹² + 78 * 0.4173² * 0.5827¹¹ + 286 * 0.4173³ * 0.5827¹⁰ ​​​

= 0.1387