1. ## find distribution

suppose that a certain device consists of 10 components and each component's life time can be modeled using a exponential distribution with mean 100 hours.

a) assuming that the device becomes out of order when one component burns. Find the distribution functionn for the lifetimee of this device.

b) assume that the device will be put to work even if at least one component is working fund the distribution functionn for the lifetime for this device.

2. part a

Quick method:
as long as the failure rates of the individual components are independant, the time to first failure is distributed as exponential with mean 10.

You can show this rigorously by noting that the exponential distribution is the waiting time of a poisson process, and that the sum of iid poisson variables is also poisson with the parameters added up.

Long Method:
Let T be the time to failure of the System
P(T<k) = P(at least 1 component dead by k)
P(T<k) = 1-P(all components alive at k)
P(T<k) = 1-(P(component 1 alive at k)*P(component 2 alive at k).....)

You can easily work out the chance of each component being alive at K, since you were given the distribution for each component.

part (b).
let T be the failure time of the system

P(T<k) = P(all components dead at time k)
P(T<k) = P(first component dead at time k) * P(second component dead at time k) * ...etc

Each term on the right hand side is the CDF of 1 component (which you know). Multiply them together to get your answer

3. Originally Posted by ilikec
suppose that a certain device consists of 10 components and each component's life time can be modeled using a exponential distribution with mean 100 hours.

a) assuming that the device becomes out of order when one component burns. Find the distribution functionn for the lifetimee of this device.

b) assume that the device will be put to work even if at least one component is working fund the distribution functionn for the lifetime for this device.

For (a) I think you need to consider the distribution of min(X1, X2, ... X10).

For (b) I think you need to consider the distribution of max(X1, X2, ... X10).

Both are simple order statistics problems and you will find the required distributions easily using Google.

4. what does find distribution function mean?

5. the distribution function gives the probability that X is less than or equal to some value.

$
F(x) = P(X \leq x)
$

7. You could try Order statistic - Wikipedia, the free encyclopedia

But, if you have not heard of order statistics before then you dont need to learn it to solve this problem. You can use the reasoning i put in my earlier reply.

Here is part (a), step by step, To get you started.

Concept - the distribution function
The distribution function of a random variable is normally written F(x). It tells you the probability that the random variable X takes a value less than or equal to some particular value (x).
$F(x) = P(X \leq x)$

Define:
$X_1 .... X_{10}$ the lifetime of the 10 components
$F_i(x) = P(X_i \leq x)$ the distribution function for each X
$T$ the lifetime of the system
$G(t) = P(T \leq t)$ the distribution function for T

Information from question
$X_i$ are independant and follow the same distribution: $exp(\lambda)$
We are told the mean value of each distribution is 100, and you should know (or be able to look up) that the mean value of an exponential distribution is $1/\lambda$
so $X_i \sim exp(0.01)$

You should know (or be able to look up) that the distribution function for he exponential distribution is
$F_i(x) = 1-e^{-\lambda x} = 1-e^{-0.01x}$

Question
You are told that the system fails when 1 component fails. Find G(t).

Step 1
$G(t) = P(T

What is the probability of the system fialing before time t? This happens if at least one component has failed by time t.
$G(t) = P(T =P(at least 1 component failes by time t)
$G(t) = P(T

Step 2
What is the probability that no component has failed by time t?
P(no failures) = $P(X_1~alive~and X_2~alive~and~X_3~alive~and...)$

P(no failures) = $P(X_1~alive)P(X_2~alive)P(X_3~alive)P(X_4~alive).. .$

P(no failures) = $P(X_1 > t)P(X_2 > t)P(X_3 > t)P(X_4>t)...$

P(no failures) = $\left(1-F_1(t) \right) \left(1-F_2(t) \right) \left(1-F_3(t) \right) \left(1-F_4(t) \right) \left(1-F_5(t)]\right)...$

P(no failures) = $[1-F(t)]^{10}$ (all X's follow the same distribution)

P(no failures) = $[1-F(t)]^{10}$

P(no failures) = $[1-1-e^{-0.01t}]^{10}$

P(no failures) = $[e^{-0.01t}]^{10} = e^{-0.1t}$

Step 3
Put it all together:

$G(t) = P(T
$G(t) = 1-e^{-0.1t}$

You can recognise this as an exponential(0.1) distribution.