Question

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.

Answer 1

`P(X=x) =\lambda^x (e^(-\lambda))/(x!), X= 0,1,2,...`

`P(X=8) = (e^(-10.4) 10.4^8)/(8!)= 0.1033`

And that would be the solution for this case.

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:

`P(X=x) =\lambda^x (e^(-\lambda))/(x!), X= 0,1,2,...`

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
`X \sim Poisson(\lambda=10.4)`

And we want to find this probability:

`P(X =8)`

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

`P(X=8) = (e^(-10.4) 10.4^8)/(8!)= 0.1033`

And that would be the solution for this case.