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.
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"
Comments
Leave a comment