Let X be the total medical expenses (in 1000s of dollars) incurredby a particula
ID: 2953975 • Letter: L
Question
Let X be the total medical expenses (in 1000s of dollars) incurredby a particular individual during a given year. Although X isa discrete random variable, suppose its distribution is quite wellapproximated by a continuous distribution with pdf f(x) = k(1+x/2.5)-7 for x0.a. What is the value of k?
b. Graph the pdf of X.
c. What are the expected value and standard deviation oftotal medical expenses?
d. This individual is covered by an insurance plan thatentails a $500 deductible provision (so the first $500 worth ofexpenses are paid by the individual). Then the plan will pay80% of any additional expenses exceeding $500, and the maximumpayment by the individual (including the deductible amount) is$2500. Let Y denote the amount of this individual's medicalexpenses paid by the insurance company. What is the expectedvalue of Y? [ Hint: First figure out what value of Xcorresponds to the maximum out-of-pocket expense of $2500. Then write an expression for Y as a function of X (which involvesseveral different pieces) and calculate the expected value of thisfunction. ]
Explanation / Answer
(b) It looks linear on [0,infinity] to me. arrange in y=mx+bform, m is your slope..etc
(c)E(x)= integral from 0 to infinity of x*f(x)dx E(x^2)= integral from 0 to infinity of x^2*f(x)dx standDev(x)= sqrt(E(x^2)-(E(x))^2)
(d) E(Y)= .8*integral over [500, 2500] of x*f(x)dx.
(b) It looks linear on [0,infinity] to me. arrange in y=mx+bform, m is your slope..etc
(c)E(x)= integral from 0 to infinity of x*f(x)dx E(x^2)= integral from 0 to infinity of x^2*f(x)dx standDev(x)= sqrt(E(x^2)-(E(x))^2)
(d) E(Y)= .8*integral over [500, 2500] of x*f(x)dx.
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