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We have a continuous random variable $X$ modelled by $f(x)=\frac{k}x$. For $1<x<9$ this is the length of time in years that water pumps last for. (Therefore $k$ can be solved to get $k=\frac{1}{2\ln3}$)

Then there are two very difficult concepts that come with this question.

First part: The farmer is offered a guarantee to cover the cost of replacing a pump that fails during the second year, at a cost of $300$. Given the pump will cost $1000$ to replace if it fails during the year, what advice would you give as to the offer?

Second part: pumps can be rented for an installation charge of $200$ plus $250$ per year payable in advance. The yearly payment is not refundable if the pump fails before the end of the year. The farmer does not purchase the guarantee. Find the probability that the pump, at the end of its life, would have cost more to rent than to buy for $1000$.

To be honest I don't even know where to start. In fact I made this account because nobody I talk to knows where to start. I hope that if somebody can explain this to me then I will be able to have a much greater understanding of the concepts underpinning this question

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For the first part, he can accept a loss of $300$ or take his chances on losing $1000$. Buying the insurance will be a good deal if the probability of the pump failing is greater than $0.3$. Integrate the probability distribution from $1$ to $2$ to get the chance it fails during the second year. For the second part, if it fails during the second year you pay $700$. If it fails during the third year you pay $950$. If it fails after the third year, you pay more than $1000$ to rent it. Compute the chance it lasts more than three years.

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  • $\begingroup$ yikes, sorry. I edited the number $k$ wrongly. $\endgroup$ – Siong Thye Goh Jul 14 '17 at 2:42
  • $\begingroup$ Thanks Ross. I completely understand how you did that. But here's another wee problem I have. I feel like there is another way I can do this problem which seems legitimate( but it can't be because it gives an answer of 0.47, whereas you method gives a perfect 0.5). The other way is this: event is Rent cost>Buy cost. Event B is always equal to 1000. Event R is modelled as 200+250X (where X is the life of the pump.) Then 200+250X>1000. I think (not sure completely) we are allowed to use algebra on this to get X>3.2. Then solving P(X>3.2) gets the answer of 0.47 $\endgroup$ – Scott Simmons Jul 14 '17 at 3:06
  • $\begingroup$ The problem says you pay 250 at the start of any year when the pump has not failed and you do not get a refund for partial years. That is why I did it the way I did. Once the fourth year starts you pay the fourth $250$ for a total $1200$, which is when the rental cost exceeds $1000$. $\endgroup$ – Ross Millikan Jul 14 '17 at 3:24
  • $\begingroup$ Wait Ross so you see how I got p(X>3.2) right? Is that a legitimate way to do the question? Then I can say that because you don't get a payment for partial years then you round it to p(X>3). Is that also a way of doing it? $\endgroup$ – Scott Simmons Jul 14 '17 at 3:48
  • $\begingroup$ I see how you got it. It is correct if you get a partial refund, but the question specifies you do not. If the pump lasts 3.1 years you spend 1200 renting and are losing compared to buying. $\endgroup$ – Ross Millikan Jul 14 '17 at 4:01

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