Question

The mills at a carpet company​ produce, on​ average, one flaw in every 500 yards of material​ produced; the carpeting is sold in​ 100-yard rolls. If the number of flaws in a roll follows a Poisson distribution and the​ quality-control department rejects any roll with two or more​ flaws, what percent of the rolls are​ rejected?

Answer 1

1.76% of the rolls are rejected.

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) = (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.

In this problem, we have that:

The mills at a carpet company​ produce, on​ average, one flaw in every 500 yards of material​ produced; the carpeting is sold in​ 100-yard rolls. This means that for each 100 yard rolls, there are expected 0.2 failures. So
`\mu = 0.2`.

If the number of flaws in a roll follows a Poisson distribution and the​ quality-control department rejects any roll with two or more​ flaws, what percent of the rolls are​ rejected?

This is
`P(X \geq 2)`.

Either the number of failures is lesser than two, or it is two or greater. The sum of the probabilities of these events is decimal 1.

`P(X < 2) + P(X \geq 2) = 1`

`P(X \geq 2) = 1 - P(X < 2)`

In which

`P(X < 2) = P(X = 0) + P(X = 1)`.

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 0) = (e^(-0.2)*(0.2)^(0))/((0)!) = 0.8187`

`P(X = 1) = (e^(-0.2)*(0.2)^(1))/((1)!) = 0.1637`

`P(X < 2) = P(X = 0) + P(X = 1) = 0.8187 + 0.1637 = 0.9824`.

Finally

`P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9824 = 0.0176`

1.76% of the rolls are rejected.