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A compressor manufacturer offers a five year warranty on repair or replacement of a compressor for its first fault. It is known that the time a compressor operates before failure is a continuous random variable $T$ with density function $$ f(t)= \begin{cases} 0, & t\le 0, \\ \frac{1}{8}e^{-t/8}, &t>0. \end{cases} $$ Calculate the average operating time of a compressor before its first fault and the probability of a compressor failing before that average time.

If the profit for the sale of a compressor is \$5,500 and its replacement or repair has a cost of \$2,000, obtain the expected profit in the sale of one of these compressors.

This is what I did:

I calculated the expected value of $T$: $$E(T)=\int_{0}^{\infty}t·\frac{e^{-\frac{t}{8}}}{8}dt=8.$$ Suppose $t$ is in years.

So, 8 years is the average time before the first fault.

Now, the probability: $$P(0\le T \le 8)=\int_{0}^{8}\frac{e^{-\frac{t}{8}}}{8}dt=1-1/e\approx 0.632$$

And I don't know how to continue from calculating the expected profit.

I think it's $$E(B)=5,500-2,000·E(T|0\le T \le 5)$$ but I don't know how to obtain $E(T|0\le T \le 5)$ and even whether it's the correct term to calculate.

gt6989b
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Aarón David Arroyo Torres
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    I think you really want $P(0\le T \le 5)$ instead of $E(T|0\le T \le 5)$ in your last formula. Are you interested into taking into account the possibility that more than one or more replacement units fail ? – WW1 Jun 26 '17 at 03:41
  • No, I don't. The exercise asks in the case of only one unit. If it's the probability $P(0\le T \le 5)$ which I have to use, how does it join to the calculation of $E(B)$? I guess I need to use a expected value to obtain $E(B)$. – Aarón David Arroyo Torres Jun 26 '17 at 09:31

1 Answers1

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The expected profit is the difference of the expected revenue and the expected costs. $E(B)=E(R)-E(C)$. And $E(R)=5,500$ So far so good. You can deal with conditional expectations.

Formally you use the law of total expectation:

$$\mathbb{E}(X)=\sum_{i=1}^n \mathbb{E}(X\mid A_{i})\cdot \mathbb{P}(A_i)$$

In your case is $n=2$. The events are $A_1=T<5$ and $A_2=T>5$. And the random variable $X$ can be replaced by the random variable for the costs, $C$. Then we have

$$\mathbb{E}(C)=\sum_{i=1}^2 \mathbb{E}(C\mid A_{i})\cdot \mathbb{P}(A_i)$$

$$\mathbb E(C)=\mathbb{E}(C\mid A_{1})\cdot \mathbb{P}(A_1)+\mathbb{E}(C\mid A_{2})\cdot \mathbb{P}(A_2)$$

$$\mathbb E(C)=\mathbb{E}(C\mid T<5)\cdot \mathbb{P}(T<5)+\mathbb{E}(C\mid T>5)\cdot \mathbb{P}(T>5)$$

It is obvious that $\mathbb{E}(C\mid T<5)=2000$. And $ \mathbb{E}(C\mid T>5)=0$. The expected replace or repair costs are $0$ since the warranty is expired after 5 five years.

callculus42
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