Question

Answer questions 16 - 17 based on the following: An insurance company has determined that each week an average of four claims are filed in their Atlanta branch.

16. What is the probability that during the next week no claims will be filed? (a) 0 (b) .0732 (c) .0498 (d) .0183 (e) 54.598

17. What is the probability that during the next week two or more claims will be filed? (a) .1464 (b) .0732 (c) .0915 (d) .8010 (e) .9085

Answer 1

4 claims are file at Atlanta branch, each week.

We model the event using the POISSION distribution, with parameter , Lambda = 4 ( mean of claims per week)

Lambda = 4 claims per week

16. P(no claims filed next week)

= e^(-lambda) * ( Lambda^x) / x! , where,

x = 0 ( no claims)

lambda = mean of events per interval = 4 in our case

= e^(-4) * ( 4^0) / 0! , where,

= 0.0183

Answers is D

17.

P(X>=2)

= 1- P(X<2)

= 1- P(X=0,1)

= 1- (e^(-4) * ( 4^0) / 0! + e^(-4) * ( 4^1) / 1!)

= 0.9085

Answer is E