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Independent Process Probability distribution

  1. Oct 18, 2015 #1
    A manufacturer has designed a process to produce pipes that are 10 feet long. The distribution of the pipe length, however, is actually Uniform on the interval 10 feet to 10.57 feet. Assume that the lengths of individual pipes produced by the process are independent. Let X and Y represent the lengths of two different pipes produced by the process.

    What is the probability that the second pipe (with length Y) is more than 0.32 feet longer than the first pipe (with length X)? Give your answer to four decimal places. Hint: Do not use calculus to get your answer.

    My work so far:

    P{(Y-X) > 0.32} = P{Y>X + 0.32}

    ∫ (from 10 to 10.57) ∫ (from x+0.32 to 10.57) (1/0.57^2) dxdy

    = 0.0614

    (It says this is the wrong answer, but i cant figure out why it is)
     
  2. jcsd
  3. Oct 18, 2015 #2

    Ray Vickson

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    Think geometrically: the "sample space" is a square of whose sides are of length 0.57. The portion of the region y > x + .32 inside the square is a triangle whose sides you can easily work out. If you know the triangle's area, how would you then get the probability?
     
  4. Oct 18, 2015 #3
    Thanks for your reply but why is there a need to find the area of a triangle?
     
  5. Oct 18, 2015 #4

    Ray Vickson

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    I cannot answer that without essentially doing your homework for you.
     
  6. Oct 18, 2015 #5
    According to my own estimation the answr is 0.582, but is late and I feel lazy. I could be wrong in my answer but I am almost sure that 0.0614 is too low for the probability of the problem. Sorry, maybe tomorrow I can come and complete the answer
     
  7. Oct 18, 2015 #6
    The answer is certainly less than 0 because the prob is less than P[X>Y] = 0.5. Indeed, the prob is less than P[X>10.32]P[Y<(10.57-.32)] < 1/4. If I were going to do this, I'd draw an XY diagram with the proper region shaded. That would show me how to set up the integral.
     
  8. Oct 19, 2015 #7
    The answer is less than 0.5 I meant to say.
     
  9. Oct 19, 2015 #8

    mathman

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    The x integral should be 10 to 10.25. I got 0.0481.
     
  10. Oct 20, 2015 #9
    you are right, but now I get 0.0962 which according to above answer is still wrong, I am not sure why
     
  11. Oct 20, 2015 #10

    mathman

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    It looks like a factor of 2. Write out the steps of your solution. It might be missing in one of your steps or I might have it, when I shouldn't.
     
  12. Oct 23, 2015 #11

    Ray Vickson

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    This answer is correct. The region {X-Y>0.32} is a triangle whose vertices are at A = (10.32,10), B = (10.57,10) and C = (10.57, 10.25), and whose area is A(triangle) = (1/2)*(.25)^2. The sample space is a square with corners at (10,10), (10,10.57), (10.57,10), 10.57, 10.57) and whose area is A(square) = (0.57)^2. The probability = p = A(triangle)/A(square) = 0.09618344106 .

    This makes sense, because the sides of the triangle are a bit less than 1/2 the sides of the square, so the area of the triangle will be a bit less than 1/8 of the square's area---that is, the probability will be a bit less than 1/8 = 0.125.
     
    Last edited: Oct 23, 2015
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