Respuesta :
Assuming a Poisson Distribution then probability it receives "k" calls is:
[tex]P(x=k) = \frac{\lambda^k e^{-\lambda}}{k!}[/tex]
where [tex]\lambda = 4, k =5[/tex]
[tex]P(x=k) = \frac{\lambda^k e^{-\lambda}}{k!}[/tex]
where [tex]\lambda = 4, k =5[/tex]
Answer: The probability it will receive 5 calls on a given that is 0.15.
Step-by-step explanation:
Since we have given that
The average number of phone inquires per day at the poison control centre = 4
So, λ = 4
Number of calls received on a given day = 5
so, k = 5
We will use "Poisson Distribution" to find the probability that it will receive 5 calls on a given day.
So, it will be written as
[tex]P(X=k)=\dfrac{\lambda^k.e^{-\lambda}}{k!}\\\\P(X=5)=\dfrac{4^5\times e^{-4}}{5!}\\\\P(X=5)=0.15[/tex]
Hence, the probability it will receive 5 calls on a given that is 0.15.