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Hopelessly Stuck!

  1. Aug 11, 2009 #1
    Hi everyone,

    I am stuck on the problem below. I think it has something to do with Poisson Process??
    I would really appreciate it if someone could point me in the right direction.

    An electronic switching device occasionally malfunctions and may need to be replaced. It is known that the device is satisfactory if it makes, on average no more than .2 errors per hour. A particular five hour period is chosen as a “test” on the device. If no more than 1 error occurs, the device is considered satisfactory. What is the probability that a satisfactory device will be mis- diagnosed as “unsatisfactory” on the basis of the test?

    Thanks
    AJ
     
  2. jcsd
  3. Aug 12, 2009 #2

    EnumaElish

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    This has to do with the Poisson Distribution. First you need to figure the mean and the variance of the distribution for the 5-hour period, then calculate the "tail" probability beyond 1.
     
  4. Aug 12, 2009 #3
    Is this correct?

    λ = 5*0.2 = 1
    P(x>1) = 1-P(x<=1) = 1-(P(x=0) + P(x=1)) = 0.2642

    Thanks
    AJ
     
  5. Aug 12, 2009 #4
    The question seems strangely worded. The problem indicates that the machine has an expected error rate of 0.2/hr. Then it asks for the probability that it will be "misdiagnosed" as unsatisfactory. The probability it is misdiagnosed may have nothing to do with whether the machine malfuctions. Bad choice of words in my opinion. I suspect what the question is getting at is if X is a random variable that counts the number of errors in the 5-hour window, then what is the probability that X is more than 1.

    This is a Poisson distributed random variable (as has been mentioned). The probability distribution function for a Poisson variable with expected rate per unit interval [itex]\lambda[/itex] is:

    [tex]P(X=n)=\frac{(\lambda t)^n e^{-\lambda t}}{n!}\text{ for }n=0,1,2,3,\dots[/tex]

    where t is the length of the interval over which X is measured.

    Here you have [itex]\lambda=0.2\text{ and }t=5[/itex] and you're looking for

    [tex]P(X>1)=1-P(x\leq1)=1-P(X=0)-P(X=1)[/tex].

    You should be able to navigate from here.

    --Elucidus
     
  6. Aug 12, 2009 #5
    Thank you very much for your help. Seems like i was on the right track.
     
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