Question

An insurance salesperson sells an average of 3 life insurance policies per week. Suppose insurance policies sales occur according to a Poisson distribution.

a) What is the probability they will sell 3 or more policies in four weeks?

b) What is the expected value of policies sold in four weeks?

c) What is the variance in the expected number of policies sold?

Answer 1

X ~ Poi ($\lambda$)

Where $\lambda$ = 3

P(X) = e^-$\lambda$ * $\lambda$ ^X / X!

a)

P(X >= 3) = 1 - P(X <= 2)

= 1 - [ P(X = 0) + P(X = 1) + P(X = 2) ]

= 1 - [ e^-3 + e^-3 * 3 + e^-3 * 3² / 2! ]

= 0.5768

b)

For 4 weeks , $\lambda$ t = 4 * 3 = 12

E(X) = $\lambda$ t = 12

c)

Var(X) =$\lambda$ = 3