Respuesta :
Answer:
32.33% probability of having at least 3 erros in an hour.
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 time interval.
The mean number of errors is 2 per hour.
This means that [tex]\mu = 2[/tex]
(a) What is the probability of having at least 3 errors in an hour?
Either you have 2 or less errors in an hour, or we have at least 3 errors. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 2) + P(X \geq 3) = 1[/tex]
We want [tex]P(X \geq 3)[/tex]
So
[tex]P(X \geq 3) = 1 - P(X \leq 2)[/tex]
In which
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-2}*(2)^{0}}{(0)!} = 0.1353[/tex]
[tex]P(X = 1) = \frac{e^{-2}*(2)^{1}}{(1)!} = 0.2707[/tex]
[tex]P(X = 2) = \frac{e^{-2}*(2)^{2}}{(2)!} = 0.2707[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1353 + 0.2707 + 0.2707 = 0.6767[/tex]
[tex]P(X \geq 3) = 1 - P(X \leq 2) = 1 - 0.6767 = 0.3233[/tex]
32.33% probability of having at least 3 erros in an hour.
