Respuesta :
Answer:
a) 0.195 = 19.5% probability that exactly three arrivals occur during a particular hour.
b) 0.762 = 76.2% probability that at least three people arrive during a particular hour
c) 2 people.
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.
Poisson process with a rate parameter of four per hour.
This means that [tex]\mu = 4[/tex].
(a) What is the probability that exactly three arrivals occur during a particular hour?
This is P(X = 3). So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 3) = \frac{e^{-4}*4^{3}}{(3)!} = 0.195[/tex]
0.195 = 19.5% probability that exactly three arrivals occur during a particular hour.
(b) What is the probability that at least three people arrive during a particular hour?
Either less than three people arrive during a particular hour, or at least three does. The sum of the probabilities of these outcomes is 1. So
[tex]P(X < 3) + P(X \geq 3) = 1[/tex]
We want [tex]P(X \geq 3) = 1 - P(X < 3)[/tex], in which
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]. So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.018[/tex]
[tex]P(X = 1) = \frac{e^{-4}*4^{1}}{(1)!} = 0.073[/tex]
[tex]P(X = 2) = \frac{e^{-4}*4^{2}}{(2)!} = 0.147[/tex]
So
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.018 + 0.073 + 0.147 = 0.238[/tex]
[tex]P(X \geq 3) = 1 - P(X < 3) = 1 - 0.238 = 0.762[/tex]
0.762 = 76.2% probability that at least three people arrive during a particular hour.
c) How many people do you expect to arrive during a 30-min period?
One hour, 4 people
Half an hour = 30 min = 4/2 = 2 people.
