Respuesta :
Answer:
A Poisson model seems reasonable for this problem, since we have the mean during the time interval.
There is a 1.9% probability that the number of autos entering the tunnel during a two-minute period exceeds three.
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 automobiles entering a mountain tunnel per two-minute period is one.
This means that [tex]\mu = 1[/tex].
For a Poisson model to be reasonable, we only need the mean during the time interval. So yes, a Poisson model seems reasonable for this problem.
Find the probability that the number of autos entering the tunnel during a two-minute period exceeds three.
We want to find [tex]P(X>3)[/tex]
Either this number is less or equal to 3, or it exceeds 3. The sum of the probabilities is decimal 1. So:
[tex]P(X \leq 3) + P(X > 3) = 1[/tex]
[tex]P(X > 3) = 1 - P(X \leq 3)[/tex]
In which
[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-1}*1^{0}}{(0)!} = 0.3679[/tex]
[tex]P(X = 1) = \frac{e^{-1}*1^{1}}{(1)!} = 0.3679[/tex]
[tex]P(X = 2) = \frac{e^{-1}*1^{2}}{(2)!} = 0.1839[/tex]
[tex]P(X = 3) = \frac{e^{-1}*1^{3}}{(3)!} = 0.0613[/tex]
So
[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.3679 + 0.3679 + 0.1839 + 0.0613 = 0.981[/tex]
Finally
[tex]P(X > 3) = 1 - P(X \leq 3) = 1 - 0.981 = 0.019[/tex]
There is a 1.9% probability that the number of autos entering the tunnel during a two-minute period exceeds three.
