Respuesta :
Answer:
a) There is a 21.69% probability of exactly 2 flaws in a 200- foot cable.
b) There is a 10.84% probability of exactly 1 flaw in the first 100 feet and exactly 1 flaw in the second 100 feet.
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
[tex]e = 2.71828[/tex] is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
a) Find the probability of exactly 2 flaws in a 200- foot cable.
Average of 0.6 per 100 feet. So 1.2 per 200 feet.
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 2) = \frac{e^{-1.2}*(1.2)^{2}}{(2)!} = 0.2169[/tex]
There is a 21.69% probability of exactly 2 flaws in a 200- foot cable.
b) Find the probability of exactly 1 flaw in the first 100 feet and exactly 1 flaw in the second 100 feet.
This is [tex]P(X = 1)^{2}[/tex]
So
[tex]P(X = 1) = \frac{e^{-0.6}*(0.6)^{1}}{(1)!} = 0.3293[/tex]
At each 100 feet, there is a 32.93% probability of exactly 1 flaw in each interval of 100 feet. So
[tex]P = (0.3293)^{2} = 0.1084[/tex]
There is a 10.84% probability of exactly 1 flaw in the first 100 feet and exactly 1 flaw in the second 100 feet.
