Respuesta :
[tex]P(k=5) = \frac{\lambda^k e^{-\lambda}}{k!} = \frac{3^5 e^{-3}}{5!} = 0.1008[/tex]
Answer:
The probability of 5 cars arriving over a five-minute interval is 0.1008 = 10.08%
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.
Mean of 3 cars arriving in five-minute time intervals.
This means that [tex]\mu = 3[/tex]
The probability of 5 cars arriving over a five-minute interval is
This is P(X = 5).
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 5) = \frac{e^{-3}*3^{5}}{(5)!} = 0.1008[/tex]
So
The probability of 5 cars arriving over a five-minute interval is 0.1008 = 10.08%