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Exponential Distribution and probability

  1. Mar 11, 2009 #1
    1. The problem statement, all variables and given/known data

    The time between calls to a corporate office is exponentially distributed with a mean of 10 minutes. What is the probability that there are more than three calls in one-half hour?

    2. Relevant equations

    F(x) = P(X <= x) = 1 - e^-(lamba*x)

    3. The attempt at a solution

    P(X > 3) = 1 - P(X < 3)
    = 1 - [1 - e^-(10*3)]
    =e^-(10*3)

    I think this answer is wrong though because I think that P(X > 3) actually means the probability that at least 30 minutes will pass before the first call instead of the probability that there are more than three calls in one half hour, but I'm not sure.
     
  2. jcsd
  3. Mar 11, 2009 #2
    P(X <= x) is not the notation for a probability distribution. It is the notation for a cumulative distribution. Additionaly I think the PDF is probably just:

    e^-(lamba*x)


    The question gave you the distribution for the average time between calls. However, you want the amount of calls that occurred in a 10 minute interval. It sounds like a bit of a conditional probability question.

    In the 10 minute intervals you can have three calls anywhere in the interval. This will be a three dimensional space of random variables a b c, where a,b,c represent when the call occurred in the interval.

    You need:
    P(a|b|c)

    I know:

    P(a|b)=P(a&b)/P(b)

    You need to find a more general expression for this. Then you need to integrate P(a|b|c) where a,b,c Go from zero minutes to 10 minutes.
     
  4. Mar 11, 2009 #3
    Looking at a solution to a similar problem in the book shows them solving it by treating it as a Poisson random variable.
     
  5. Mar 11, 2009 #4

    Mark44

    Staff: Mentor

    P(X > 3) = 1 - P(0 <= X <= 3) = 1 - P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
    Now, all you need is calculate the four probabilities on the right, add them, and subtract the total from 1.

    Mark
     
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