Respuesta :
Answer:
1.76% of the rolls are rejected.
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:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] 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 [tex]\mu = 0.2[/tex].
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 [tex]P(X \geq 2)[/tex].
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.
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex].
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-0.2}*(0.2)^{0}}{(0)!} = 0.8187[/tex]
[tex]P(X = 1) = \frac{e^{-0.2}*(0.2)^{1}}{(1)!} = 0.1637[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.8187 + 0.1637 = 0.9824[/tex].
Finally
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9824 = 0.0176[/tex]
1.76% of the rolls are rejected.
