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| Re: last min help |
2004-03-01 21:45:00 <Avraham Adler> |
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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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