Question

The number of surface flaws in plastic panels used in the interior of automobiles has a...

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.07 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.

(c) If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws? Round your answers to four decimal places (e.g. 98.7654).

Homework Answers

Answer #1

For a car, lets find a probability of flaw
Here, λ = 0.07*10 = 0.7 and x = 0
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X > 0) = 1 - P(X <= 0).
P(X > 0) = 1 - (0.7^0 * e^-0.7/0!)
P(X > 0) = 1 - (0.4966)
P(X > 0) = 1 - 0.4966 = 0.5034

Now,
Here, n = 10, p = 0.5034, (1 - p) = 0.4966 and x = 1
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 1).
P(X <= 1) = (10C0 * 0.5034^0 * 0.4966^10) + (10C1 * 0.5034^1 * 0.4966^9)
P(X <= 1) = 0.0009 + 0.0092
P(X <= 1) = 0.0101

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