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Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 4 - Problem 26e
Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 4 - Problem 26e

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Let X be the total medical expenses (in 1000s of ISBN: 9780321629111 32

Solution for problem 26E Chapter 4

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

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Problem 26E

Let ?X ?be the total medical expenses (in 1000s of dollars)incurred by a particular individual during a given year. Although ?X ?is a discrete random variable, suppose its distribution is quite well approximated by a continuous distribution with pdf f(x) = k ( 1+ x/2.5)–7for x?0 . a.? hat is the value of ?k?? b.? ?Graph the pdf of ?X?. c.? ?What are the expected value and standard deviation of total medical expenses? d.? ?This individual is covered by an insurance plan that entails a $500 deductible provision (so the first$500 worth of expenses are paid by the individual). Then the plan will pay 80% of any additional expenses exceeding $500, and the maximum payment by the individual (including the deductible amount) is$2500. Let ?Y ?denote the amount of this individual’s medical expenses paid by the insurance company. What is the expected value of ?Y?? [?Hint?: First figure out what value of ?X ?corresponds to the maximum out-of-pocket expense of $2500. Then write an expression for ?Y ?as a function of ?X ?(which involves several different pieces) and calculate the expected value of this function.] Step-by-Step Solution: Step 1 of 3 Problem 26E Answer: Step1: We have Let X be the total medical expenses (in 1000s of dollars)incurred by a particular individual during a given year.Although X is a discrete random variable, suppose its distribution is quite well approximated by a continuous distribution with pdf f(x) = k 7 , x0 . (1+2.5 We need to find, a. What is the value of k b. Graph the pdf of X. c. What are the expected value and standard deviation of total medical expenses d. This individual is covered by an insurance plan that entails a$500 deductible provision (so the first $500 worth of expenses are paid by the individual). Then the plan will pay 80% of any additional expenses exceeding$500, and the maximum payment by the individual (including the deductible amount) is $2500. Let Y denote the amount of this individual’s medical expenses paid by the insurance company. What is the expected value of Y Step2: a). Consider, x 7 F (x) = k(1 + 2.5) Integrate above equation with respect to x then we get by taking limits from 1 t. P(X 0) = f(x)dx = k1 + x ) dx 0 1 2.5 7 = k [(0.4x + 1) ]dx 1 Let u = 0.4x + 1 du = 2.5dx Substitute in above integral we get 7 = k [u] 0.4du 1 u71 = 2.5k [ 71 1 6 = 2.5k [ ]6 1 6 = 2.5k [ ]u 6 1 (0.4x+1) = 2.5k [ 8 ]1 = 2.5k/6 6 k = 2.5 Therefore, the value of k is 2.4. b). When k = 2.4 the graph of the given pdf is given below Step3: c). Consider, x 7 F (x) = k(1 + 2.5) Now, x 7 E(X) = xf(x)dx = kx1 + 2.5) dx 1 1 = k [x(0.4x + 1) ]dx 1 Let u = 1+ 2.5 du = 0.4dx Substitute in above integral we get 2.5(2.5u2.5) = k u7 du 1 (2.5u2.5) = 2.5k u7 du 1 7 = 2.5k (2.5u-2.5) u (2.5) k = 30 3 6 = (2.5)2.5 30 = 1 2 Therefore, E(X) = 0.5. similarly, 7 E(X ) = x f(x)dx = kx(1 +2 x ) dx 1 1 2.5 3 = (2.5) k 60 (2.5)2.5 = 60 5 = 8 = 0.6250 Therefore,E(X ) = 0.6250. Hence, 1 Mean() = 2 and Variance( ) = E(X ) - [E(X)] 2 = 0.6250 - [0.5]2 = 0.375 Standard deviation( ) = 0.375 = 0.6123. Therefore, standard deviation is 0.6123. Step4: d). Let h(x) is zero if x 500,the deductible amount thereafter it pays 0.8 of the cost over 500, namely 0.8(x-500) until the customer paid$ 2500.If z is the amount where customer has paid exactly $2500, then 2500 = x (x 500) 5 Which means z = 10500.hence the pmf is given by Then the expected payout is x 7 E(h(x)) = h(x)k(1 + 2.5 0 10500 = k 4 (x 500)(1 + x ) dx+ k (x 2500)(1 + x ) dx 5 2.5 2.5 500 10500 4201 = 5k (2.5u 502.5)u du+k (2.5u 2502.5)u du7 201 4201 12 2 1 1 402 1 1 12 1 2502.5 = 5 { 5 2015 5) 6 (2016 6)+ 5 {(2•42015 6•4201 )}. 4201 4201 Step 2 of 3 Chapter 4, Problem 26E is Solved Step 3 of 3 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 The answer to “Let ?X ?be the total medical expenses (in 1000s of dollars)incurred by a particular individual during a given year. Although ?X ?is a discrete random variable, suppose its distribution is quite well approximated by a continuous distribution with pdf f(x) = k ( 1+ x/2.5)–7for x?0 . a.? hat is the value of ?k?? b.? ?Graph the pdf of ?X?. c.? ?What are the expected value and standard deviation of total medical expenses? d.? ?This individual is covered by an insurance plan that entails a$500 deductible provision (so the first $500 worth of expenses are paid by the individual). Then the plan will pay 80% of any additional expenses exceeding$500, and the maximum payment by the individual (including the deductible amount) is $2500. Let ?Y ?denote the amount of this individual’s medical expenses paid by the insurance company. What is the expected value of ?Y?? [?Hint?: First figure out what value of ?X ?corresponds to the maximum out-of-pocket expense of$2500. Then write an expression for ?Y ?as a function of ?X ?(which involves several different pieces) and calculate the expected value of this function.]” is broken down into a number of easy to follow steps, and 187 words. This textbook survival guide was created for the textbook: Probability and Statistics for Engineers and the Scientists, edition: 9. This full solution covers the following key subjects: individual, Expenses, Expected, medical, Insurance. This expansive textbook survival guide covers 18 chapters, and 1582 solutions. Probability and Statistics for Engineers and the Scientists was written by and is associated to the ISBN: 9780321629111. Since the solution to 26E from 4 chapter was answered, more than 635 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 26E from chapter: 4 was answered by , our top Statistics solution expert on 05/06/17, 06:21PM.

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