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Probability w/ continuous R.V

  1. Jan 2, 2014 #1
    1. The problem statement, all variables and given/known data

    A commercial water distributor supplies an office with gallons of water once
    a week. Suppose that the weekly supplies in tens of gallons is a random
    variable with pdf



    f(x) = 5(1-x)^4 , 0 < x <1

    f(x) = 0 , elsewhere

    Approx how many gallons should be delivered in one week so that the probability of the supply is 0.1?
    2. Relevant equations



    3. The attempt at a solution

    I started out by finding the cdf


    F(y) = 0 , y < 0

    F(y) = -(1-y)^5 + 1 , 0 ≤ y < 1

    F(y) = 1 , y ≥ 1



    Here is where i get lost. So y is our gallons (in tens). We want to find the amount of gallons that yields a probability of 0.1

    p(y≤t) = F(t) = 0.1 thus

    -(1-y)^5 + 1 = 0.1


    When I solve for y I do not get the right answer which is apparently 4 gallons
     
  2. jcsd
  3. Jan 2, 2014 #2

    vela

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    Is that exact wording of the problem statement as given to you?
     
  4. Jan 2, 2014 #3

    haruspex

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    Clearly the probability of any specific quantity is zero, so is the question asking for a y such that the probability of at least y is .1 or the probability of at most y is .1? you have interpreted it the second way.
     
  5. Jan 3, 2014 #4

    HallsofIvy

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    I am not clear on what "the probability of the supply is 0.1" means. I suspect you mean "the probability that the amount actually supplied is at least y is 0.1". That means that you want to find y such that [tex]\int_0^y 5(1- x)^4 dx[/tex].

    To integrate [itex]\int_0^y 5(1- x)^4 dx= 0.1[/itex] let u= 1- x so that du= -dx, when x= 0, u= 1, and when x= y, u= 1- y so that the integral becomes [itex]5\int_1^{1- y} u^4(-du)= \int_{1- y}^1 u^4 du[/itex]
     
  6. Jan 3, 2014 #5

    Ray Vickson

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    He did that integration already in his OP. The real issue seems to be that he/she does not know whether or not to solve F(y) = 0.1 or 1-F(y) = 0.1.
     
  7. Jan 3, 2014 #6
    Hello, DotKite.

    You did the integration correctly. The problem is that you solved for the variable incorrectly.

    By definition of probability of continuous variable, if $F(y)$ is cdf, we mean:

    ##F(y) = P(Y \leq y)##

    With a lower bound ##a##, ##P(a \leq Y \leq y) = F(y) - F(a)##

    You also wrote out the expression correctly, but the wrong value. Try to solve for ##y## again, and you should get the correct answer.
     
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