An insurance salesman sells on average 10.4 life insurance policies per week. Find the probability that in a particular week he sells 8 policies. Round your answer to 4 decimal places.

Respuesta :

Answer:

[tex]P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}, X= 0,1,2,...[/tex]

[tex] P(X=8) = \frac{e^{-10.4} 10.4^8}{8!}= 0.1033[/tex]

And that would be the solution for this case.

Step-by-step explanation:

Previous concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

[tex]P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}, X= 0,1,2,...[/tex]

Solution to the problem

Let X the random variable that represent the number of life insurance policies that the salesman person sells. We know that [tex]X \sim Poisson(\lambda=10.4)[/tex]

And we want to find this probability:

[tex] P(X =8)[/tex]

And we can use the probability mass function and we got:

[tex] P(X=8) = \frac{e^{-10.4} 10.4^8}{8!}= 0.1033[/tex]

And that would be the solution for this case.