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

Answer:

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

$P(X=8) = \frac{e^{-10.4} 10.4^8}{8!}= 0.1033$

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:

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

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:

$P(X=8) = \frac{e^{-10.4} 10.4^8}{8!}= 0.1033$

And that would be the solution for this case.