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Exercise 30 (Section 3.3) gave the pmf of Y, the number of

Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye ISBN: 9780321629111 32

Solution for problem 53E Chapter 3

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 53E

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.

Step-by-Step Solution:

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. Step 2 : b). Now we have to calculate fewer than half have at least one citation. P(x) = 1-p(citation) P(x) = 1- 0.60 P(x) = 0.40 This is a little different than part a. Now the success probability is 0.40 for having at least one citation. Let Y be the number of person with at least one citation. So Y ~ Binomial (15,0.40) P( Y < 15/2 ) = P( Y < 7.5 ) 7 y 15y P(Y < 7.5) = (15 )(0.4y) (1 0.40) y=0 0 150 1 151 2 152 P(Y < 7.5) = (15 )(0040) (1 0.40) + (15 1(0.40) (1 0.40) + (15 2(0.40) (1 0.40) + (15 )(0.40) (1 0.40) 153 + 3 (15 )(0.40) (1 0.40) 154 + (15 )(0.40) (1 0.40) 155+ 4 6 5 156 7 157 (15 6(0.40) (1 0.40) + (15 7(0.40) (1 0.40) P(Y < 7.5) = 0.00047 + 0.004702 + 0.021942 + 0.063388 + 0.126776 + 0.185938 + 0.206598 + 0.177084 P(Y < 7.5) = 0.786897 0.7869 P(Y < 7.5) = 0.7869 Therefore fewer than half have at least one citation is 0.769 Step 6 : c). Here we have to calculate the number that have at least one citation is between 5 and 10, inclusive. 10 y 15y P(5 Y 10) = (15 )(0.4y) (1 0.40) y=5 5 155 6 156 7 157 P(5 Y 10) = (15 )(0.50) (1 0.40) + (156)(0.40) (1 0.40) + (15 7(0.40) (1 0.40) + (15 8(0.40) (1 0.40) 158 + (15 9(0.40) (1 0.40) 159+ (15 100.40) (1 0.40) 1510 P(5 Y 10) = 0.1859 + 0.2065 + 0.1770 + 0.1180 + 0.0612 + 0.0244 P(5 Y 10) = 0.77337 0.7734 P(5 Y 10) = 0.7734 Therefore the number that have at least one citation is between 5 and 10 is 0.7734.

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Chapter 3, Problem 53E is Solved
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Textbook: Probability and Statistics for Engineers and the Scientists
Edition: 9
Author: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye
ISBN: 9780321629111

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Exercise 30 (Section 3.3) gave the pmf of Y, the number of

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