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Math Help - Probability of a Compenent failing

  1. #1
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    Probability of a Component failing

    A critical component on a submarine has an operating lifetime that is exponentially distributed with a mean of 6 months. As soon as a component fails, it is replaced bya new one having statistically identical properties. What is the minimum number ofspare components which the submarine ought to carry if it is leaving for a one yeartour of duty and it is desired that the probability of having an inoperable unit causedby failures exceeding the spare inventory be less than 0.02

    Let V represent the critical component

    E(V) = 0.5 => \lambda = 2

    The probability of the first time the component fails is:

    P(V \ fails) = 2e^{-2}

    Kind of stuck here. Any hints/suggestions would be greately appreciated.
    Last edited by statmajor; March 14th 2011 at 02:58 PM.
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  2. #2
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    How about this: Suppose X_1, ..., X_n are iid and distributed according to \mbox{Exp}(1). Then

    \displaystyle<br />
\sum_{i = 1} ^ n X_i \sim \mbox{Gamma}(n, 1).
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  3. #3
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    So \displaystyle<br />
\sum_{i = 1} ^ n V_i \sim \mbox{Gamma}(n, 2)

    EDIT: Okay, I think I got it

    I need to find the probability that

    So \displaystyle<br />
P(\sum_{i = 1} ^ n X_i <1)<0.02

    E(\Sigma X_i) = 0.5n Var(\Sigma X_i)=0.25n

    Applying the central limit theorem:

    P(Z<\frac{1-0.5n}{0.25\sqrt{n}})<0.02

    \frac{1-0.5n}{0.25\sqrt{n}}=2.05

    Solving for n, I get n=0.9835

    So they need to bring 2 components.

    Have I made a mistake anywhere?
    Last edited by statmajor; March 14th 2011 at 06:25 PM.
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  4. #4
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    I'm not as bold as some, but I'm pretty sure using the CLT for n = 2 isn't a good idea.

    Gammas are related to chisquare random variables, so you have tables that give you the quantiles for them.
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  5. #5
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    Quote Originally Posted by theodds View Post
    I'm not as bold as some, but I'm pretty sure using the CLT for n = 2 isn't a good idea
    I'm not sure what you meant by that. I didn't use the CLT for n = 2. I used to to solve for n and got n =2
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  6. #6
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    But the CLT isn't, typically, a good approximation for n = 2. The reasoning behind what you did only applies when the n that you get is sufficiently large the the CLT to hold.

    Your reasoning is essentially:

    (1) The n I need is large, so apply the CLT
    (2) Solve for n
    (3) Get n = 2

    You see the problem, yes?
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  7. #7
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    Quote Originally Posted by theodds View Post
    Gammas are related to chisquare random variables, so you have tables that give you the quantiles for them.
    A Chi-Square is a special case of the gamma where \Gamma (n,2)~Chi-Square where n is the degrees of freedom. But I still don't know what the degrees of freedom are.
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  8. #8
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    Quote Originally Posted by statmajor View Post
    A Chi-Square is a special case of the gamma where \Gamma (n,2)~Chi-Square where n is the degrees of freedom. But I still don't know what the degrees of freedom are.
    Actually, it is the same as a chisquare with 2n degrees of freedom (not n).

    You are supposed to choose n, or equivalently, the degrees of freedom of an appropriate chi-square. Here's one more hint:

    \displaystyle<br />
Z \sim \mbox{Gamma}(\alpha, \beta) \iff \frac{2Z} \beta \sim \chi^2 _{2\alpha}<br />
    where we are using the shape/scale parameterization of the Gamma (i.e. \mbox E Z = \alpha\beta). You can look up in a table an appropriate cutoff to get the n you want.
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