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Homework Help: Probability Density

  1. Nov 12, 2012 #1
    The manager of a fast food restaurant determines that the average time that her customers wait for their food is 2.5 minutes. The manager wants to advertise that anybody who isn't served within a certain number of minutes gets a free hamburger. She doesn't want to give away free hamburgers to more than 2% of her customers. What should the advertisement say?

    I'm solving for [itex]t[/itex]:

    [itex]\int_ 0.4e^{-t/2.5}~dt=0.02[/itex]

    [itex]-e^{-t/2.5}=0.02[/itex]

    [itex]0=0.02+e^{-t/2.5}[/itex]

    Take the natural logs and add [itex]-t/2.5[/itex] to get it back on the other side.

    [itex]t/2.5=-3.91[/itex]

    [itex]t=-1.56[/itex]

    The answer is ten minutes. What did I do wrong?
     
    Last edited: Nov 12, 2012
  2. jcsd
  3. Nov 12, 2012 #2

    Ray Vickson

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    Homework Helper

    You cannot have exp(-t/2.5) = -0.02, since the exponential function is always > 0. You did the integration incorrectly.

    RGV
     
  4. Nov 12, 2012 #3

    ehild

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    You need to show the limits of the integral and calculate with them. The probability that somebody is not served for x minutes is

    [tex]\int _x^\infty{0.4 e^{-0.4 t}dt}=0.02[/tex]

    ehild
     
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