Question

In a certain automobile manufacturing paint shop, paint defects on the hood occur at a mean rate of 0.8 defect per square meter. A hood on a certain car has an area of 3 square meters. (a) Justify the use of the Poisson model. (b) If a customer inspects a hood at random, what is the probability that there will be no defects? (c) One defect? (d) Fewer than two defects?

Answer 1

Considering the situation described, we have that:

a) The Poisson distribution is used because we only have the mean.

b) There is a 0.0907 = 9.07% probability that there will be no defects.

c) There is a 0.2177 = 21.77% probability that there will be one defect.

d) There is a 0.3084 = 30.84% probability that there will be fewer than two defects.

What is the Poisson distribution?

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

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

The parameters are:

• x is the number of successes

• e = 2.71828 is the Euler number

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

Item a:

The Poisson distribution is used because we only have the mean.

We have 0.8 defects per m² and 3 m², hence the mean is given by:

$\mu = 3 \times 0.8 = 2.4$

Item b:

The probability is P(X = 0), hence:

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

$P(X = 0) = \frac{e^{-2.4}(2.4)^{0}}{(0)!} = 0.0907$

There is a 0.0907 = 9.07% probability that there will be no defects.

Item c:

The probability is P(X = 1), hence:

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

$P(X = 1) = \frac{e^{-2.4}(2.4)^{1}}{(1)!} = 0.2177$

There is a 0.2177 = 21.77% probability that there will be one defect.

Item d:

The probability is P(X < 2), hence:

P(X < 2) = P(X = 0) + P(X = 1) = 0.0907 + 0.2177 = 0.3084.

There is a 0.3084 = 30.84% probability that there will be fewer than two defects.

More can be learned about the Poisson distribution at https://brainly.com/question/13971530