Respuesta :
Answer:
0.1535 = 15.35% probability that at least one customer arrives in a particular one minute period.
0.1443 = 14.43% probability that at least two customers arrive in a particular 4 minute period.
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.
Find the probability that at least one customer arrives in a particular one minute period.
The average is of 10 customers per hour(60 minutes), so for this case, [tex]\mu = \frac{10}{60} = \frac{1}{6}[/tex]
This is:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-\frac{1}{6}}*(\frac{1}{6})^{0}}{(0)!} = 0.8465[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.8465 = 0.1535[/tex]
0.1535 = 15.35% probability that at least one customer arrives in a particular one minute period.
Probability that at least two customers arrive in a particular 4 minute period.
The average is of 10 customers per hour(60 minutes), so for this case, [tex]\mu = \frac{10*4}{60} = \frac{4}{6} = \frac{2}{3}[/tex]
This probability is:
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-\frac{2}{3}}*(\frac{2}{3})^{0}}{(0)!} = 0.5134[/tex]
[tex]P(X = 1) = \frac{e^{-\frac{2}{3}}*(\frac{2}{3})^{1}}{(1)!} = 0.3423[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.5134 + 0.3423 = 0.8557[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.8557 = 0.1443[/tex]
0.1443 = 14.43% probability that at least two customers arrive in a particular 4 minute period.
