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Probability distribution, find constant

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data

    x = 0, P(x) = 0.4
    x = 1, P(x) = 0.1
    x = 2, P(x) = 0.1
    x = 3, P(x) = 0.1
    x = 4, P(x) = 0.3

    If P(x)=k(5-x) for x = 0,1,2,3,4, find value of constant k

    3. The attempt at a solution

    0.4 = k(5-0)
    0.1 = k(5-1)
    0.1 = k(5-2)
    0.1 = k(5-3)
    0.3 = k(5-4)

    5k+4k+3k+2k+k=1
    15k=1
    k = 1/15
     
  2. jcsd
  3. Jun 26, 2011 #2

    LCKurtz

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    Are you sure you have stated the problem correctly? The sum of the probabilities is already 1.
     
  4. Jun 26, 2011 #3

    HallsofIvy

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    The probability distribution,
    x = 0, P(x) = 0.4
    x = 1, P(x) = 0.1
    x = 2, P(x) = 0.1
    x = 3, P(x) = 0.1
    x = 4, P(x) = 0.3

    does NOT satisfy P(x)= k(5- x). If it did, then you would have to have P(0)= 0.4= k(5)so that k= 0.4/5= 0.08 but then P(1)= k(5-1)= 0.08(4)= 0.32, not 0.1. If P(x)= k(5- x) then 15k= 1 because any probability distribution must sum to 1, not because of "0.4+ 0.1+ 0.1+ 0.1+ 0.3= 1".
     
  5. Jun 26, 2011 #4

    LCKurtz

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    Deathfish, I have apparently misunderstood your problem. Tell me, are the numbers 0.4,0.1,0.1,0.1, and 0.3 you have listed above supposed to have been somehow given in the statement of the problem or they a result of your attempt at solving the problem? If they are results of your work they shouldn't be stated as part of the problem. If they are the result of your work, they are wrong as Halls has pointed out.
     
  6. Jun 27, 2011 #5
    0.4,0.1,0.1,0.1, and 0.3 are the values of P(x) in the question... values of x and corresponding P(x) are listed down in a table although i dont know how to post a table here... i have no idea what the question means by "If P(x)=k(5-x) for x = 0,1,2,3,4, find value of constant k" this i am copying down from the question too
     
  7. Jun 27, 2011 #6

    HallsofIvy

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    If P(x)= k(5- x), then it can not be the same P as for the given list.
     
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