Probability of a Compenent failing

**statmajor** · March 14th 2011, 10:44 AM

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.

**theodds** · March 14th 2011, 01:09 PM

How about this: Suppose $X_1, ..., X_n$ are iid and distributed according to $\mbox{Exp}(1)$ . Then

$\displaystyle \sum_{i = 1} ^ n X_i \sim \mbox{Gamma}(n, 1)$ .

**statmajor** · March 14th 2011, 01:35 PM

So $\displaystyle \sum_{i = 1} ^ n V_i \sim \mbox{Gamma}(n, 2)$

EDIT: Okay, I think I got it

I need to find the probability that

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

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

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?

**theodds** · March 15th 2011, 07:44 AM

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.

**statmajor** · March 15th 2011, 08:53 AM

Originally Posted by theodds

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

**theodds** · March 15th 2011, 08:56 AM

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?

**statmajor** · March 15th 2011, 09:12 AM

Originally Posted by theodds

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.

**theodds** · March 15th 2011, 10:10 AM

Originally Posted by statmajor

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 Z \sim \mbox{Gamma}(\alpha, \beta) \iff \frac{2Z} \beta \sim \chi^2 _{2\alpha} $
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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