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Exponential distribution, two exercises

  1. Jul 14, 2013 #1
    1. The problem statement, all variables and given/known data
    Waiting time in a restaurant is exponentially distributed variable, with average of 4 minutes. What is the probability, that a student will in at least 4 out of 6 days get his meal in less than 3 minutes?


    2. Relevant equations



    3. The attempt at a solution
    If I understand correctly I could say that ##f(t)=e^{-\lambda t}##, meaning ##P(t<3)=\int_{0}^{3}\omega (t)dt=\int_{0}^{3}(\frac{f(t)}{dt})dt=\int_{0}^{3}\lambda e^{-\lambda t}dt##

    Ok? This is now probability that a student will get his meal in less than 3 minutes, so the probability that it would take longer is ##1-P(t<3)## and the final result should be something like ##\sum_{i=4}^{6}\begin{pmatrix}
    6\\
    i
    \end{pmatrix}p^{i}(1-p)^{6-i}##

    Does this sound right? How do I determine ##\lambda##???

    1. The problem statement, all variables and given/known data
    Elapsed time until a device suddenly stops working is exponentially distributed with median 4h. Calculate the probability, that the device will work at least for 5 hours!



    2. Relevant equations



    3. The attempt at a solution
    Very different as before, but still have no idea how to get ##\lambda##???
     
  2. jcsd
  3. Jul 14, 2013 #2

    Ray Vickson

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    Your notation is unfortunate; the usual expression for the density of the exponential is
    [tex]f(t) = \left\{ \begin{array}{l} \lambda e^{-\lambda t},\; t \geq 0\\
    f(t) = 0, \;t < 0 \end{array} \right. [/tex] The (cumulative) distribution function is
    [tex] F(t) = \int_0^t f(x) \, dx = 1 - e^{-\lambda t},[/tex]
    while the complementatry cumulative is ##P\{ T > t \} = 1 - F(t) = e^{-\lambda t}.##

    You get the mean by using standard formulas found in any textbook or on-line. Alternatively, you can perform the integration ##E T = \int_0^{\infty} t f(t) \, dt## and see what you get.

    BTW: the easiest way to get a binomial coefficient in TeX is to use the command {n \choose m}, which gives ##{n \choose m}.##
     
    Last edited: Jul 14, 2013
  4. Jul 14, 2013 #3
    That is what I had in my head. =)

    Hmmm... Just cheking:##F(t)=\int_{0}^{t }f(t)dt=\int_{0}^{t }e^{-\lambda t}dt=-\frac{1}{\lambda }\int_{0}^{u_1 }e^{u}du=\frac{1}{\lambda }(1-e^{-\lambda t})##
    What does ##\frac{1}{\lambda }## tell me, if anything, because I don't see it in your calculation?

    AHA! So ##E(t)=\int_{0}^{\infty }tf(t)dt=\int_{0}^{\infty }te^{-\lambda t}dt=\frac{\lambda t+1}{\lambda ^{2}}=4## Which gives me a quadratic equation for ##\lambda## where the right soultion is with + so ##\lambda =\frac{1+\sqrt{5}}{8}##


    Thanks for helping and this hint!
     
  5. Jul 14, 2013 #4

    Ray Vickson

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    The density should read as ##f(t) = \lambda e^{-\lambda t}##; I have edited the post to correct this. So, subsequent expressions are OK: ##F(t) = 1 - e^{-\lambda t}##, etc. Think of it this way: if t has dimensions of time, λ must have dimensions 1/time = rate per unit time; that makes the product λt dimensionless, which it must be if we are going to exponentiate it. Similarly, probabilities are dimensionless numbers, so F(t) cannot have λ or 1/λ outside the exponential.

    Your computation of ET is incorrect; getting λ is much, much easier than what you wrote. Look in a book, or look on line!
     
  6. Jul 14, 2013 #5
    Ammm, this is written in my notebook, just as it is written on wikipedia: http://en.wikipedia.org/wiki/Expected_value#Univariate_continuous_random_variable

    So expected value ##E(t)=\int_{-\infty }^{\infty }tf(t)dt## but ##t## can only be positive and ##f(t)=\lambda e^{-\lambda t}##.

    I am a bit confused at this moment? How is this incorrect?
     
  7. Jul 14, 2013 #6

    Ray Vickson

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    This part is correct, but you are not finished. You need to do the integral (or consult a book or a web page, as I have said now for the third---and last---time). In a previous post you gave some weird expression for the mean that was incorrect (for one thing, it involved 't, but that has already been integrated out). You had a quadratic equation to solve, which is totally false.
     
  8. Jul 14, 2013 #7
    According to wolfram alpha and my own calculations with the help of a book the integral is http://www.wolframalpha.com/input/?i=axe^(-ax)+integrate

    ##E(t)=\int_{0}^{\infty }t\lambda e^{-\lambda t}dt=-\frac{1}{\lambda }(e^{-\infty}(\lambda t+1))+\frac{1}{\lambda }(e^{0}(\lambda t+1))=\frac{1}{\lambda }(\lambda t+1)=4##
    now ##4\lambda=\lambda t+1##

    What is wrong here?
     
  9. Jul 14, 2013 #8

    Ray Vickson

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    There should be no 't' in the final answer. As I said before, 't' has been integrated out---it is gone.
     
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