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Uniform distribution

  1. May 4, 2013 #1

    the answer is 2/3:

    my attempt:
    one side X~U[1,7]

    the longer part of this side, call Y, where Y~[4,7]

    P(Y>6) = 1/3

    don't see how they got 2/3, they have 3 different methods in the answers, but none doing my method.

    I'm thinking, as there are two sides in the rectangle and I've only limited mine to one side then if I multiply that by 2 I'll get the answer - or is that even relevant?

    I'll give another example where I used my method and it works:


    part c)

    the shorter side can be distributed by X~U[0,10]

    P(X>8) = 1/5 which is the answer, but this doesn't work for the original method. Any ideas why? Thanks.
  2. jcsd
  3. May 4, 2013 #2


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    I agree that X ~ U[1,7].

    Where are you getting that Y ~ U[4,7]?

    I would think that instead Y~U[3,9], where Y is the length of the other side. If one side of the rectangle is 1 cm, then the other side must be 9 cm. Similarly, if one side of the rectangle is 7 cm, the other side must be 3 cm. So Y ~ U[3,9]. Do you see where to go from here?

    Is this an Edexcel S2 question?
  4. May 4, 2013 #3
    Yes it is

    I got Y ~ U[4,7] where Y is the longer side, by setting Y as the longer side (not the other side). Median length is 4, so for Y to be the longer side, it has to be between [4,7].

    I get the correct answer Y ~ U[3,9] where Y is the other side, and doing P(X>6) + P(Y>6), but I don't see why it doesn't work using the above method.
  5. May 4, 2013 #4


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    I don't follow your reasoning. Your distribution doesn't represent Y to be the longer side. If Y ~ U[4,7], what happens when 5 < X < 7? Clearly, X is then the longer side.
  6. May 4, 2013 #5
    hm, I'm just confused now.

    If they are asking for one side, and if Y ~ U[3,9] then why do we have to consider the other side (i.e X). Why can't we just consider Y
  7. May 4, 2013 #6


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    They are asking what happens when the longer side is greater than 6 cm.

    If X ~ U[1,7] and Y~U[3,9], then either side can be longer than 6 cm. We can have X > 6, or Y > 6. We can't just consider Y, because X can be larger than 6 cm also. Does that make sense?
  8. May 4, 2013 #7
    that makes sense

    but if you see in the second example, I split up the whole length and I just considered the shorter side (which is what I tried to do when doing the mock), but it didn't work :\.
  9. May 4, 2013 #8


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    You mean this?

    I agree with the reasoning above. The shorter side must be between 0 and 10, otherwise it isn't the shorter side by definition.
  10. May 4, 2013 #9

    Ray Vickson

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    For the first question above: let Y be the OTHER (not longest) side and let Z = max(X,Y) be the length of the longest side. We have X+Y=10, so Y = 10 - X. We have Z = X if X > Y = 10-X, so Z = X if X > 5. Of course, Z = 10-X if X < 5. Draw a graph of the function Z(x) = max(x,10-x) for 4 <= x <= 7. What is the smallest value of Z(x)? What is the largest value? For what x-region do we have Z(x) > 6? What is the probability of that x-region?

    BTW: I agree with your answer: P{Z > 6} = 1/3.
    Last edited: May 4, 2013
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