1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted