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The time between process problems in a manufacturing line is exponentially distributed with a mean of 30 days.

(a) Let T be the waiting time (in days) for four problems. What is the distribution of T?

(b) What is the expected waiting time for four problems?

(c) What is the probability that the time until the fourth problem exceeds 120 days?

(d) Suppose that the problems can be classified into three mutually exclusive classes: I, II and III. The probability that a problem is of type I, II, and III, are respectively, 0.75, 0.2, 0.05.

(i) Give the distribution of the waiting time for a process problem of type I.

(ii) Suppose that the problems I, II and III, cost (per problem): 1,000, 2,500 and 6,000 $, respectively. Give the mean total cost for process problems for a period of 90 days.

(e) There are no process problems for 30 days, what is the probability that there will be at least one process problem in the next 30 days?


For (a) (b) and (c) I have managed to answer the following:

(a) The waiting time distribution for T is Erlang since this is a sum of exponential distributions.

(b) E[X] = r.1/lambda = 4/lambda , where lambda = 1/30 ==> 4/0.033

(c) This would be a Poisson distribution P(X>120)

However for (d) and (e) I am kind of stuck. Any help would be greatly appreciated!

Thank you,

Ali.

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1 Answer 1

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For (e), the answer lies is the memorylessness of the exponential. However long we have waited, the probability of at least one problem in the next $30$ days is the same as the unconditional probability of at least one problem, for an exponential of mean $30$. This is $1-e^{-30/30}$.

For (d)(i), we are dealing with an exponential with mean $30/(0.75)=40$.

For (c), you have given an outline of an approach with no detail. We want the probability that $X\le 3$, where $X$ is a certain random variable with Poisson distribution whose parameter I expect you can identify.

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For (c) we want the 4th problem to exceed 120 days, so 121st day or more. Since from (b) we have already calculated the expected time for the fourth problem to be 4/0.033 which is equal to 121.21. Can we just say P(X<=4) with parameter 1/30? –  Alistair Feb 7 '12 at 8:34
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We are dealing with a continuous distribution, so we exceed $120$ days even if we do it by a microsecond, and $121$ has nothing much to do with it. Your $4/(0.033)$ is the product of a rounding error, you should instead have written in the solution of (b) $4/(1/30)=120$. And for (c), we really want $P(X\le 3)$, and the Poisson has parameter $120(1/30)=4$. So the probability is $e^{-4}(1+4+4^2/2!+4^3/3!)$. –  André Nicolas Feb 7 '12 at 8:56
    
Thanks Andre, this really helped me put things in focus. For (c) P(X>120) = P(X<=3) = e^(-4)*(1+4+4^2/2!+4^3/3!). I should start working with factors rather than rounded numbers to get proper results as with the case for my lambda. Thanks again! –  Alistair Feb 7 '12 at 9:20
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@Alistair: Ultimately, we have to feed expressions into a calculator/computer to get answers. I like to do the feeding late, not at every step. That way, one retains feeling for the structure of the formulas. Also, it is good to use the "memory" feature of the calculator. Saves time, helps prevent roundoff errors. –  André Nicolas Feb 7 '12 at 14:58
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@Alistair: First line in your comment, you write $P(X>120)=P(X\le 3)$. On looking at it, you can see that makes no sense. You probably know what you meant by that. Let $Y$ be the random variable that measures the waiting time until the $4$-th problem. Let $X$ be rv that counts the number of problems in $120$ days. Then $P(Y>120)=P(X\le 3)$. –  André Nicolas Feb 7 '12 at 15:05

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