Calculating the Expected Value of an Appliance Warranty

In summary: 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.
  • #1
Hodgey8806
145
3

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!
 
Physics news on Phys.org
  • #2


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
 
  • #3


Would you care to share how you calculated the Expected value of v(t)?
 
  • #4
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
 

1. What is the formula for calculating the expected value of an appliance warranty?

The formula for calculating the expected value of an appliance warranty is: (probability of a defective appliance) x (cost of repair or replacement) - (cost of warranty). This formula takes into account the likelihood of a defect, the potential cost of repair or replacement, and the cost of the warranty itself.

2. How is the probability of a defective appliance determined?

The probability of a defective appliance is determined by analyzing historical data, such as the percentage of appliances that have needed repair or replacement in the past. It can also be estimated by considering the quality and reliability of the manufacturer and the specific appliance in question.

3. Is it always more cost effective to purchase an appliance warranty?

No, it is not always more cost effective to purchase an appliance warranty. It ultimately depends on the cost of the warranty and the likelihood of a defect. If the cost of the warranty is high and the probability of a defect is low, it may not be worth the investment.

4. Can the expected value of an appliance warranty change over time?

Yes, the expected value of an appliance warranty can change over time. As an appliance ages, the likelihood of a defect may increase, making the warranty more valuable. Additionally, the cost of repair or replacement may also change, impacting the overall expected value.

5. How can the expected value of an appliance warranty be used in decision making?

The expected value of an appliance warranty can be used to make an informed decision about whether or not to purchase the warranty. If the expected value is positive, it may be worth purchasing the warranty. If the expected value is negative, it may be more cost effective to forgo the warranty and pay for repairs or replacement as needed.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
280
  • Differential Equations
Replies
0
Views
293
  • Calculus and Beyond Homework Help
Replies
13
Views
11K
  • Calculus and Beyond Homework Help
Replies
10
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
492
  • Calculus and Beyond Homework Help
Replies
8
Views
620
  • Calculus and Beyond Homework Help
Replies
7
Views
684
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
Back
Top