Statistics Problems and Solutions Second Edition 2B.9

Question

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So I was (again) looking at Statistics Problems and Solutions (Second Edition) by Bassett et al., 1986 when I saw this question on page $97$:

$2$B.$9\quad$ Guaranteed life of a machine

The lifetime in hours of a certain component of a machine has the continuous probability density function$$f(x)=\frac{1}{1000}e^{-x/1000},x\geq0$$The machine contains five similar components, the lifetime of each having the above distribution. The makers are considering offering a guarantee that not more than two of the original components will have to be replaced during the first $1000$ hours of use. Find the probability that such a guarantee would be violated, assuming the components wear out independently, and that if a component does fail then the replacement used of particularly high quality and will certainly last for the $1000$ hours.

My attempt on this

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1. We know that all $5$ components of the machine have the same continuous probability density function$$f(x)=\frac{1}{1000}e^{-x/1000},x\geq0$$
2. We know that if the components wear out, it will be replaced with a component of higher quality that will $100$% of the time last for the full $1000$ hours of use, however, we do not need to know this to solve the question.

My steps that I took

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$$\operatorname{Pr}(\text{lifetime}\lt1000\text{hours})=\int_{0}^{1000}\frac{1}{1000}e^{-x/1000}\frac{d}{dx}$$$$\iff e^{-(0)/1000}-e^{-(1000)/1000}$$$$\iff(1)-e^{(-1)}$$$$\iff x\approx0.6321$$Now, we can find the required probability by binomial distribution:$$\operatorname{Pr}(X=3)+\operatorname{Pr}(X=4)+\operatorname{Pr}(X=5)$$$$=\frac{\left(\frac{5!}{3!}\right)}{1!2!}(0.6321)^3(0.3679)^2+5(0.6321)^4(0.3679)+(0.6321)^5$$ $$ \begin{align} \approx73.6403\text{%} \end{align} $$ Therefore, it would be unwise to offer the guarantee, due to the high chance that $2$ or more parts are going to malfunction.

To clarify

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1. This is different from my question here.

My question

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Is my solution correct, or is there another way that I could rewrite my thought process to make it correct?

Answer 1

Yes, your solution is correct, although it is a little strange that you wrote $$\frac{\left(\frac{5!}{3!}\right)}{1! \, 2!}$$ rather than simply $$\binom{5}{3} = \frac{5!}{3! \, 2!}.$$

To further your understanding:

1. The original question made a claim of $1000$ hours. What would be the maximum time $t$ that the manufacturer can claim such that the guarantee that no more than $2$ components will fail within $t$ hours will be true with at least $95\%$ probability?
2. The original question used $n = 5$ components. What would be the minimum number of components $n$ such that a guarantee at least $3$ original components will still be operational after $1000$ hours will be true with at least $95\%$ probability?

Answers:

$t \le 209.802$ hours

$n \ge 15$ components