Respuesta :
Answer:
a) There is a 16.0623% probability that 5 messages are received in 1 hour.
b) There is a 11.5880% probability that 10 messages are received in 1.5 hours.
c) There is a 22.4042% probability that 2 messages are received in 0.5 hours.
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 time interval.
In this problem, we have a mean of 6 messages per hour.
(a) What is the probability that 5 messages are received in 1 hour?
Find the value of P when [tex]x = 5[/tex] and [tex]\mu = 6[/tex]
So
[tex]P(X = 5) = \frac{e^{-6}*(6)^{5}}{(5)!} = 0.160623[/tex]
There is a 16.0623% probability that 5 messages are received in 1 hour.
(b) What is the probability that 10 messages are received in 1.5 hours?
The mean is 6 messages in one hour.
For 1.5 hours, the mean is 6*1.5 = 9 messages.
So
We have to find the value of P when [tex]x = 10[/tex] and [tex]\mu = 9[/tex].
[tex]P(X = 10) = \frac{e^{-9}*(9)^{10}}{(10)!} = 0.115880[/tex]
There is a 11.5880% probability that 10 messages are received in 1.5 hours.
(c) What is the probability that less than 2 messages are received in 1/2 hour?
The mean is 6 messages in one hour.
For 0.5 hours, the mean is 6*0.5 = 3 messages.
So
We have to find the value of P when [tex]x = 2[/tex] and [tex]\mu = 3[/tex].
[tex]P(X = 10) = \frac{e^{-3}*(3)^{2}}{(2)!} = 0.224042[/tex]
There is a 22.4042% probability that 2 messages are received in 0.5 hours.
