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.]

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