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Continuous probability

  1. Oct 11, 2015 #1
    1. The problem statement, all variables and given/known data

    The length of satisfactory service (years) provided by a certain model of laptop computer is a random variable having the probability density
    f(x) = (1/4.5)e^(-x/4.5) for x > 0
    and 0 for x <= 0

    find the probabilities that one of these laptops will provide satisfactory service for
    a) at most 2.5 years
    b) anywhere from 4 to 6 years
    c) at least 6.75 years

    3. The attempt at a solution
    first we evaluate the integral

    ∫(1/4.5)e^(-x/4.5)dx = (1/4.5) ∫ e^(-x/4.5)dx = (1/4.5)(-4.5)e^(-x/4.5) = -e^(-x/4.5)

    part a) evaluate from 0 to 2.5

    -e^(-2.5/4.5) - (-e^0) = 1 - e^(-2.5/4.5) = .4262

    b) evaluate from 4 to 6
    -(e^(-6/4.5) - e^(-4/4.5)) = .1475

    c) from 6.75 to ∞
    -(e^(-∞) - e^(-6.75/4.5)) = -(0 - .2231) = .2231

    is my reasoning correct?
     
  2. jcsd
  3. Oct 11, 2015 #2

    Orodruin

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    Yes. It is correct.
     
  4. Oct 11, 2015 #3
    thank you
     
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