A life insurance salesman sells on the average 4 life insurance policies per week. a. What is the probability that in a given week he will sell three policies? (round your answer to three decimal places)
b. What is the probability that he will sell at most 5 policies in two weeks period? ( round your answer to three decimal places)

Suppose x be a random variable that refers to the number of life insurance plans sold by salespeople using Poisson. Assuming that the average number of insurance policies sold per week is

we know that,

For point a:

Probability which is required:  

For point b:

For the next 2 weeks, the average amount of insurance policies sold:

Probability which is required:  

where

Similarly,

Therefore,  

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0.195

0.191

Step-by-step explanation:

Given that :

μ = 4

Using poisson :

P(x) = (e^-μ * μ^x) ÷ x!

P(x = 3) = (e^-4 * 4^3) / 3!

= (0.0183156 * 64) / 6

= 0.1953664

= 0.195

b. What is the probability that he will sell at most 5 policies in two weeks period?

Mean per week = 4

Mean in two weeks = 4 * 2 = 8

P(x ≤ 5) = p(5) + p(4) +... + p(0)

Using the poisson distribution calculator :

P(x ≤ 5) = 0.19124

= 0.191

80%

step-by-step explanation:

step-by-step explanation: