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Here is the actual question: $A$ is random variable representing the lifespan of a component. It is an exponential law with an average of 10. Considering a system with $n$ components $A$, what is the minimum value of $n$ so that the system has a fiability of 0.999 for a 4 year period?

I thought about approximating it with the normal law (even if it's probably a pretty bad approximation). This gave me this random variable $Z_n = (Y-\nu)/(\sigma \sqrt{n})$.

Then I figured I was looking for $\mathbb P(Y \geq 4) = 1 - \mathbb P(Y \leq 3)$. In this case $\mathbb P(Y \leq 3)$ would be 0.001 since $\mathbb P(Y \geq 4) = 0.999$. With the cdf table I find that it gives -3.09. So I solve the equation $(3-10n) / (10 \sqrt{n}) = -3.09$. This gives an answer of 10 (around that). The actual answer is 7.

What have I done wrong and how could I find the answer of that question?

Edit : The components are independent from each other; if only one works, the system works. So basically only one of the components needs to have a life span of 4 years. The problem doesn't mention if the components are replaced so I assumed they weren't.

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    This question does not seem well-posed as of yet. When does the system fail? When all $n$ components have failed where they all start functioning from the beginning? Or, are the $n$ components replaced one after another once the previous one fails? The answer you're looking for appears to depend on such details.2011-10-30
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    The components are independent from each other; if only one works, the system works. So basically only one of the components needs to have a life span of 4 years. The problem doesn't mention if the components are replaced so I assumed they weren't.2011-10-30
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    Then, you're looking for the maximum. Here is a **hint**: $$\mathbb P(\text{system lasts $x$ years}) = 1 - \mathbb P(\text{system fails before $x$ years})$$ and $$\mathbb P(\text{system fails before $x$ years}) = \mathbb P(A_1 < x, A_2 < x, \ldots, A_n < x) \>.$$ Can you continue from there?2011-10-30
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    I think I already had figured out the first one. As for the second one, I'm not sure what this represents. The individual probability that each component fails before x years?2011-10-30
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    If the system fails before $x$ years, then that means that *all* of the components failed before $x$ years. (Right?) And, the latter is the *joint* probability that each component has a lifetime of less than $x$. Now, to continue, think about the extra structure you have in your problems. Namely, can you rewrite the right-hand side using assumptions about the $A_i$?2011-10-30

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If (and this is a big if) what you mean is that the $n$ components all start at time zero, that each component has a lifetime exponentially distributed with mean $10=1/\lambda$ and if the question is to know for which minimal value of $n$ one has $p_n\leqslant\varepsilon$ where $\varepsilon=1/1'000$ and $p_n$ is the probability that none of the $n$ components is still in function at time $t=4$, then here is your answer.

The probability that a given component is not functioning at time $t$ is $p_1=1-\exp(-\lambda t)$ hence the probability that none of them is functioning is $p_n=(p_1)^n=(1-\exp(-\lambda t))^n$. Numerically, $p_6=0.0013$ and $p_7=0.0004$ hence $p_6>\varepsilon>p_7$ and the minimal $n$ is $7$.

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    Oh, and there is no approximation by a normal law in the picture (this is not needed and not asked and anyway, $n=6$ or $7$ would be rather small values for this approach).2011-10-30
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    Thanks a lot for the answer! I thought it was an approximation by normal law because this was the subject of the chapter. Seems like it was much simpler than that.2011-10-30
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    I'm just wondering, though, isn't the probability $(1-\lambda\exp(-\lambda t))$?2011-10-30
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    No it is not, see [here](http://en.wikipedia.org/wiki/Exponential_distribution).2011-10-30
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For a continuous random variable:

$$P(Y\le 3) \neq P(Y<4) \; ,$$

therefore use

$$P(Y\ge 4) = 1 - P(Y < 4) \; .$$

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    Why the restriction to continuous random variables?2011-10-30
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    Because $Y$ is a continuous random variable in his problem. And you'll notice that he applies an identity that only holds for discrete variables. That's why his end result is wrong.2011-10-30
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    My point is *your* identity has no such restriction! :) (I think the OP's issues start even before that.)2011-10-30
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    OK, I agree that it is true for discrete variables as well, but I meant to say that $P(Y\le 3) \neq P(Y<4)$ for continuous random variables.2011-10-30
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    I tried solving with what you proposed and it still doesn't give me the correct answer. Is my method of using the approximation with the normal distribution wrong?2011-10-30
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    Yes, that is also incorrect, since your $Y$ variable as you constructed it is a kind of average lifetime, but what is needed is looking at the maximum lifetime. See Didier's answer.2011-10-30