Question

A life insurance salesman works on commission. Suppose the sales, on average, 3 policies per week.

a) What is the probability that he fails to sell any policies in a week? b) What is the probability that he sells between 3 and 5 policies in one week. Note that for Poisson distributions, the mean is proportional to the length of the interval. c) Given that a work-week for the salesman is five days, what is the probability that the insurance salesman sells exactly one policy on any given workday?

- Please use Poisson distribution

Answer 1

Probability mass function of a Poisson distribution with mean :

X : Number of policies salesman sells in a week

X follows a Poisson distribution with =3

(a) Probability that he fails to sell any policies in a week = P(X=0)

Probability that he fails to sell any policies in a week = 0.049787068

(b) Probability that he sells between 3 and 5 policies in one week = P(3X5) = P(X=3)+P(X=4)+P(X=5)

P(3X5) = P(X=3)+P(X=4)+P(X=5) = 0.224041808+0.168031356+0.100818813=0.492891977

Probability that he sells between 3 and 5 policies in one week = 0.492891977

(c)

Number of days in a work-week =5

Average number of policies sold in a workday = Average number of policies sold in a workweek/Number of days in a work-week =3/5 = 0.6

Y : Number of policies salesman sells in a workday

Y follows Poisson distribution with mean =0.6

Probability that the insurance salesman sells exactly one policy on any given workday = P(Y=1)

Probability that the insurance salesman sells exactly one policy on any given workday =0.329286982