Question

A life insurance salesman sells on the average 3 life insurance policies per week, assuming Poisson distribution is followed. Compute the probability he sells more than 2 or more policies for duration of 2 weeks.

Answer 1

We have given that,

$\mu = 3$

We have to find $P(X \ge 2)$

Using the Poisson distribution,

$P(X) = \dfrac{e^{-\mu}\mu^x}{ x!}$

then,

$P(X \ge 2) = 1-(P(0)+P(1))$

$= 1- \dfrac{e^{-3}}{1} - \dfrac{e^{-3}(3)}{1}$

$= 0.811$