Respuesta :
Answer:
[tex] P(X=6)[/tex]
And using the probability mass function we got:
[tex]P(X=6) =3^6 \frac{e^{-3}}{6!}=0.05041[/tex]
So then we can conclude that the statement is True.
Step-by-step explanation:
Definitions and concepts
The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:
[tex]P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}[/tex]
And the parameter [tex]\lambda= 3[/tex] represent the average ocurrence rate per unit of time.
Solution to the problem
For this case we want to find this probability:
[tex] P(X=6)[/tex]
And using the probability mass function we got:
[tex]P(X=6) =3^6 \frac{e^{-3}}{6!}=0.05041[/tex]
So then we can conclude that the statement is True.
Answer:
The given statement is True.
Step-by-step explanation:
We are given that the number of airplanes arriving at an airport per minute is a Poisson process with the mean number of airplanes arriving per minute is 3.
Let X = Distribution of number of airplanes arriving at an airport per minute
So, X ~ Poisson([tex]\lambda[/tex])
The mean of Poisson distribution is given by, E(X) = [tex]\lambda[/tex] = 3
which means, X ~ Poisson(3)
The probability distribution function of a Poisson random variable is:
[tex]P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!}; for x=0,1,2,3...[/tex]
Probability that exactly 6 planes arrive in the next minute = P(X = 6)
P(X = 6) = [tex]\frac{e^{-3}*3^{6}}{6!}[/tex] = [tex]\frac{e^{-3}*729}{720}[/tex] = 0.05041
Therefore, the given statement is true.
