Calculating the Expected Value of an Appliance Warranty

Hodgey8806
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Homework Statement


The initial value of an appliance is $700 and it's value in the future is given by:

v(t)=100(2^(3-t)-1), 0<=t<=3

where t is time in years. Thus, after the first 3 years the appliance is worth nothing as far as the warranty is concerned. If it fails in the first three years, the warranty pays v(t). Compute the expected value of the payment on the warranty if T has an exponential distribution with mean 5.


Homework Equations



I would think to use E(x) = int(x*v(t), 0<=t<=3) which is approximately 574.14977

I would also use the probability of it failing in the first three years which would be:
P(T<=3) = 1- e^-(3/5) which is approximation .4512 .

The Attempt at a Solution



Thus, my solution would be .4512*574.14977 which is approximately $259.05.

Is this correct? Thanks!
 
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Hodgey8806 said:

Homework Statement


The initial value of an appliance is $700 and it's value in the future is given by:

v(t)=100(2^(3-t)-1), 0<=t<=3

where t is time in years. Thus, after the first 3 years the appliance is worth nothing as far as the warranty is concerned. If it fails in the first three years, the warranty pays v(t). Compute the expected value of the payment on the warranty if T has an exponential distribution with mean 5.


Homework Equations



I would think to use E(x) = int(x*v(t), 0<=t<=3) which is approximately 574.14977

I would also use the probability of it failing in the first three years which would be:
P(T<=3) = 1- e^-(3/5) which is approximation .4512 .

The Attempt at a Solution



Thus, my solution would be .4512*574.14977 which is approximately $259.05.

Is this correct? Thanks!

I get $121.73, and my calculation is nothing like yours. I am computing the expected value of v(T); I have no idea what you are computing.

RGV
 


Would you care to share how you calculated the Expected value of v(t)?
 
Hodgey8806 said:
Would you care to share how you calculated the Expected value of v(t)?
Sorry, no. I am allowed to give hints only, and I have already said exactly what I did: I computed the expectation of a function of a random variable.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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