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Central limit theorem [probability]

  1. Apr 5, 2012 #1
    1. In any one-minute interval, the number of requests for a popular Web page is a Poisson random variable with expected value 360 requests.

    A Web server has a capacity of C requests per minute. If the number of requests in a one-minute interval is greater than C the server is overloaded. Use the central limit theorem to estimate the smallest value of C for which the probability of overload is less than 0.025.

    Because it's a Poisson distribution then E[X] = 360 = alpha = Var[X]
    I'm using a Z table, so at 0.5-0.025 = 0.475, Z = 1.96
    so Phi((x-360/sqrt(360)) = 1.96
    and I get x = 397.1884 which is wrong.

    am I on the right track, or completely off?
     
  2. jcsd
  3. Apr 5, 2012 #2

    Mark44

    Staff: Mentor

    I don't see anything glaringly wrong. What are you given as the right answer?
     
  4. Apr 5, 2012 #3
    I'm not given a correct answer, my webwork just tells me if my answer is right or wrong and I get a certain number of tries.
     
  5. Apr 5, 2012 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    Your solution method is OK. Your answer is only a very little bit wrong. The exact answer, using the Poisson and solving numerically, is C = 398.1530823, which we should round up to 399. You should round yours up to 398.

    RGV
     
  6. Apr 5, 2012 #5
    Ah okay. Thank you!
     
  7. Apr 5, 2012 #6

    Ray Vickson

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    Homework Helper

    Actually, your solution is more-or-less exact; in my previous response I mistakenly used C-1 instead of C. The precise solution, using the exact Poisson distribution, is C = 397.15308; this expresses P{X > C} = P{X >= C+1} in terms of incomplete Gamma functions, and so makes sense even for non-integer C. So, the usable solution is to round up to 398. Your normal approximation of 397.1884 is very close tot the exact, and rounds up to exactly the same integer 398.

    RGV
     
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