Solution Found!
Exercise 30 (Section 3.3) gave the pmf of Y, the number of
Chapter 4, Problem 53E(choose chapter or problem)
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.
Questions & Answers
QUESTION:
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: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.