# Approximating a sum of exponential distribution with a normal distribution

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. –  cardinal Oct 30 '11 at 18:55
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. –  Djeezus Oct 30 '11 at 18:57
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? –  cardinal Oct 30 '11 at 19:01
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? –  Djeezus Oct 30 '11 at 19:07
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 $A_i$? –  cardinal Oct 30 '11 at 19:11

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). –  Did Oct 30 '11 at 19:17
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. –  Djeezus Oct 30 '11 at 19:19
I'm just wondering, though, isn't the probability $(1-\lambda\exp(-\lambda t))$? –  Djeezus Oct 30 '11 at 19:36
No it is not, see here. –  Did Oct 30 '11 at 19:39

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? –  cardinal Oct 30 '11 at 19:04
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. –  Raskolnikov Oct 30 '11 at 19:06
My point is your identity has no such restriction! :) (I think the OP's issues start even before that.) –  cardinal Oct 30 '11 at 19:07
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. –  Raskolnikov Oct 30 '11 at 19:08
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? –  Djeezus Oct 30 '11 at 19:11