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 last min help 2004-03-01 21:45:00
 Hi, Some short and quick questions: 1) Exam Nov. 2001 #22 "A dose of 250 of a certain drug is injected into a patient everyday at noon. The amount of the drug that remains in the body from each injection is given by r(t) = 250e^(-t/6), where t is the time in days since the injection. Calculate the lesat upper bound for how much the drug will be in a patient's body if the injections are given indefinitely. Solution: 250/(1-e^(-1/6)). Would you please explain why is (-1/6)? thanks 2) Exam May 2000 #18 An insurance policy reimburses dental expense, X, up to a maximum benefit of 250. Te probability density funtion for X is f(x) = ce^(-0.004x) for x>=0 and 0, otherwise, where c is a constant. How to solve c? I tried to use intergral but I got c = -0.004. 3) Exam Nov 2000 #25 A manufacturer's annual looses follow a distribution with density function f(x) = (2.5(0.6)^2.5)/x^3.5 for x>0.6. To cover its losses, the manufacturer purchases an insurance policy with an annual deductible of 2. How can you tell this is a policy limit question? Thanks in advance! KS _________________________________________________________________ MSN Photos is the easiest way to share and print your photos: http://photos.msn.com/support/worldwide.aspx

 Re: last min help 2004-03-01 21:45:00
 At 02:39 AM 5/21/02 -0400, you wrote: >Hi, > >Some short and quick questions: > >1) Exam Nov. 2001 #22 > "A dose of 250 of a certain drug is injected into a patient everyday > at noon. The amount of the drug that remains in the body from each > injection is given by r(t) = 250e^(-t/6), where t is the time in days > since the injection. Calculate the lesat upper bound for how much the > drug will be in a patient's body if the injections are given indefinitely. > > Solution: 250/(1-e^(-1/6)). Would you please explain why is (-1/6)? >thanks There is 250*e^-(0/6) immediately after todays injection (t=0) There is 250*e^-(1/6) from yesterdays injection (t=1) There is 250*e^-(2/6) from two days ago injection (t=2) . . There is 250*e^-(n/6) from n days ago There is 250* Sum {i= 1,2,3...n,n+1,..+inf} [e^(-1/6)]^n This is the geometric series with p = e^(-1/6). Sum of infinite geometric series with |p| < 1 is 1/1-p So here we get 250 * (1/(1-e^-1/6)) >2) Exam May 2000 #18 > > An insurance policy reimburses dental expense, X, up to a maximum > benefit of 250. Te probability density funtion for X is >f(x) = ce^(-0.004x) for x>=0 and 0, otherwise, where c is a constant. >How to solve c? I tried to use intergral but I got c = -0.004. Firstly, just recognize that this is an exponential with ß = .004 so se= MUST be .004 Or INT (0,inf) ce^-.004x dx = 1 c * -(1/.004)e^-.004x | (0,inf) = 1 c/.004 = 1 c= .004 Don't forget that INT e^-ax = e^-ax/(-a) >3) Exam Nov 2000 #25 > > A manufacturer's annual looses follow a distribution with density > function f(x) = (2.5(0.6)^2.5)/x^3.5 for x>0.6. To cover its losses,= the > manufacturer purchases an insurance policy with an annual deductible of 2. >How can you tell this is a policy limit question? > >Thanks in advance! The losses not paid by the insurance company here (b/c of the deductible) can also be considered a "self-insurance" policy by the company where it pays itself and has a limit of two. Does that make sense? It barely does to= me :). --Avraham

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