Respuesta :
Answer:
0.735 = 73.5% probability of getting at least 4 calls between eight and nine in the morning.
Step-by-step explanation:
We have the mean during a time interval, which means that the Poisson distribution is used 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 interval.
The number of calls received by an office on Monday morning between 8:00 AM and 9:00 AM has a mean of 5.
This means that [tex]\mu = 5[/tex]
Calculate the probability of getting at least 4 calls between eight and nine in the morning.
This is:
[tex]P(X \geq 4) = 1 - P(X < 4)[/tex]
In which
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-5}*5^{0}}{(0)!} = 0.0067[/tex]
[tex]P(X = 1) = \frac{e^{-5}*5^{1}}{(1)!} = 0.0337[/tex]
[tex]P(X = 2) = \frac{e^{-5}*5^{2}}{(2)!} = 0.0842[/tex]
[tex]P(X = 3) = \frac{e^{-5}*5^{3}}{(3)!} = 0.1404[/tex]
Then
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0067 + 0.0337 + 0.0842 + 0.1404 = 0.265[/tex]
[tex]P(X \geq 4) = 1 - P(X < 4) = 1 - 0.265 = 0.735[/tex]
0.735 = 73.5% probability of getting at least 4 calls between eight and nine in the morning.
