Calculating the Expected Value of an Appliance Warranty

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Homework Help Overview

The problem involves calculating the expected value of a warranty payment for an appliance, given its initial value and a future value function over a specified time period. The time until failure is modeled with an exponential distribution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the expected value calculation using the function v(t) and the probability of failure within the first three years. There are differing interpretations of the expected value computation, with one participant questioning the method used by another.

Discussion Status

The discussion is ongoing, with participants sharing their calculations and questioning each other's approaches. Some guidance is offered in the form of hints about computing expectations, but no consensus has been reached regarding the correct method or outcome.

Contextual Notes

Participants are working under the constraints of homework rules, which limit the type of assistance that can be provided. There is also a focus on the assumptions related to the distribution of failure times and the value function.

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
 

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