## The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.06 flaws per s

Question

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.06 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel. (a) What is the probability that there are no surface flaws in an auto’s interior

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3 months 2022-02-02T12:42:17+00:00 1 Answer 0 views 0

There is a 54.88% probability that there are no surface flaws in an auto’s interior.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$$

In which

x is the number of sucesses

$$e = 2.71828$$ is the Euler number

$$\mu$$ is the mean in the given interval.

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.06 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.

For one square foot, we have 0.06 flaws.

So for 10 square feet, the mean is $$\mu = 10*0.06 = 0.6$$

(a) What is the probability that there are no surface flaws in an auto’s interior

This is P(X = 0).

$$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$$

$$P(X = 0) = \frac{e^{-0.6}*(0.6)^{0}}{(0.6)!} = 0.5488$$

There is a 54.88% probability that there are no surface flaws in an auto’s interior.