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Exponential distribution problem

  1. Oct 14, 2015 #1
    1. The problem statement, all variables and given/known data
    The amount of time that a surveillance camera will run without having to be reset is a random variable having the exponential distribution with beta = 50 days. find the probabilities that such a camera will
    a) have to be reset in less than 20 days
    b) not have to be reset in at least 60 days


    3. The attempt at a solution
    first we integrate the function
    f(x) = (1/beta)e^(-x/beta) = (1/50)e^(-x/50)
    after integrating we get
    -e^(-x/50) evaluated from a to b
    for part a)
    we are evaluating from 0 to 20 so we get
    -(e^(-20/50) - 1) = 1 - e^(-20/50) = .3297

    for part b) we evaluate from 60 to ∞
    -(0 - e^(-60/50)) = e^(-60/50) = .3012

    is this method correct?
     
  2. jcsd
  3. Oct 14, 2015 #2

    andrewkirk

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    Gold Member

    It looks right to me.
     
  4. Oct 14, 2015 #3

    Mark44

    Staff: Mentor

    It looks OK to me as well.
    Another way to do part b is this: ##Pr(x \ge 60) = 1 - Pr(0 \le x < 60)##
    So ##\frac 1 {50}\int_{60} ^{\infty} e^{-x/50}dx = \frac 1 {50}\left(1 - \int_0^{60}e^{-x/50}dx\right)##
     
  5. Oct 17, 2015 #4
    ok great thank you!
     
  6. Oct 17, 2015 #5

    Ray Vickson

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    Yes.

    Here are a small number of handy facts about the exponential distribution---worth committing to memory (and pretty easy to derive for yourself).

    If ##X## has distribution ##\text{Exp}(\lambda)## (so that its pdf on ##\{ x > 0 \}## is ##\lambda e^{-\lambda x}##) then:
    [tex] \begin{array}{cl}
    1)& EX = \displaystyle \frac{1}{\lambda} \\
    2)&\text{Var} X = \displaystyle \frac{1}{\lambda^2} \\
    3)& \text{Coefficient of variation} \equiv \displaystyle \frac{\text{standard deviation}}{\text{mean}} = 1 \\
    4) & P(X > x) = e^{- \lambda x} , \; x \geq 0
    \end{array}
    [/tex]
    From 4) it follows that ##P(X \leq x) = 1 - e^{- \lambda x}## for ##x \geq 0##. Also, remembering 3) is a convenient way of remembering 2).
     
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