Question

Question 4. (15 points) An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of moving violations for which the individual was cited during the last 3 years. The pmf of Y is 0 p(y) 0.25 0.100.05 (1). Find a, such that p(y) is a p.m.f (2). Write out the c.d.f of completely. (3) Suppose an individual with Y violations incurs a surcharge of S (r +100Y'. Calculate the expected amount of the surcharge.

Answer 1

a) Total probability is 1, so

$a+0.25+0.10+0.05=1\\ \Rightarrow a=0.60$

b) The CDF is defined as $P(y)=\Pr(Y\leq y)$

y	0	1	2	3
P(y)	0.60	0.85	0.95	1

c) The expected value is given by

$\left E(Y+100Y^2)=(0+100\times 0^2)(0.60)+ (1+100\times 1^2)(0.25)+ (2+100\times 2^2)(0.10)+(3+100\times 3^2)(0.05) \right \\ ~~~~~~~~~~~~~~~~~~~~~~~~~=110.60$