Respuesta :
Answer:
40.82% probability that 2 or more calls arrive in any 2 minute period.
Step-by-step explanation:
We have the mean during a time interval, so we use the poisson distribution to solve this question.
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.
10 minutes.
3 outside calls, 4 inside. So mean of 3+4 = 7.
That is, 0.7 calls per minute.
Compute the probability that 2 or more calls arrive in any 2 minute period.
2 minutes, so [tex]\mu = 2*0.7 = 1.4[/tex]
Either less than two calls arrive, or two or more do. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]. So
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-1.4}*(1.4)^{0}}{(0)!} = 0.2466[/tex]
[tex]P(X = 1) = \frac{e^{-1.4}*(1.4)^{1}}{(1)!} = 0.3452[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.2466 + 0.3452 = 0.5918[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5918 = 0.4082[/tex]
40.82% probability that 2 or more calls arrive in any 2 minute period.
