Exercise 30 (Section 3.3) gave the pmf of ?Y, ?the number of traffic citations for a randomly selected individual insured by a particular company. What is the probability that among 15 randomly chosen such individuals a.?? t least 10 have no citations? b.? ?Fewer than half have at least one citation? c.? ?The number that have at least one citation is between 5 and 10, inclusive?* Reference exercise 30 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 a.? ?Compute ?E?? ?). b. ?Suppose an individual with ?Y ?violations incurs a surcharge of $100?Y?2. Calculate the expected amount of the surcharge.

Answer : Step 1 : Let Y be the number of moving violations for which the individual was cited during the last 3 years. The pmf of Y is y p(y) 0 0.60 1 0.25 2 0.10 3 0.05 The pmf of Y, the number of traffic citations for a randomly selected individual insured by a particular company. Now we have to calculate the probability that among 15 randomly chosen such individuals. a). At least 10 have no citations. Let X be the number of person with no citation. So X ~ Binomial (15,0.60) The formula of the binomial distribution is n n x nx P(X) = ( )(px (1 p) x=0 15 15 x 15x P(X 10) = ( )(0.x0) (1 0.60) x=10 P(X 10) = ( 15 )(0.60) (1 0.60) 1510 + (15 ) (0.60) (1 0.60) 1511+ ( 15)(0.60) (1 0.60) 1512+ 10 11 12 ( 15) (0.60) (1 0.60) 1513 +( 15 )(0.60) (1 0.60) 1514+ 15 1513 1515 14 ( 15 ) (0.60) (1 0.60) ) P(X 10) = 0.185938 + 0.126776 + 0.063388 + 0.021942 + 0.004702 + 0.00047 P(X 10) = 0.403216 Therefore at least 10 have no citations is 0.403216.