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Discrete probability distribution

  1. May 13, 2008 #1
    1. The problem statement, all variables and given/known data
    The problem is as shown in the attatchment.


    2. Relevant equations
    The relevant equations are also given in the attatchment.


    3. The attempt at a solution
    My problem is how to adapt the given formula in order to find the sum of the function k(40-r)

    Do i use the formula for 1 to 40, then 1 to 20 and subtract?

    Even just a hint in the right direction would be useful. This sorta stuff has never been my strong point.
     

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  3. May 13, 2008 #2
    Think i might have got it.

    k(1[tex]\rightarrow[/tex]40[tex]\sum(40-r)[/tex] - 1[tex]\rightarrow[/tex]19[tex]\sum(40-r)[/tex]) + 20k = 1

    k(40*40 - [tex]\frac{40(41)}{2}[/tex] - (19*20 - [tex]\frac{19(20)}{2}[/tex])) = 1

    230k =1
    k = 1/230


    EDIT sorry its a bit messy, dont know how to notate the limits for the summation properly.
     
    Last edited: May 13, 2008
  4. May 14, 2008 #3
    How did

    [tex]\sum_{r=1}^{19}(40-r) = (19)(20) - \frac{(19)(20)}{2}[/tex]

    The part I don't understand is the (19)(20). And what happened to the 20k? Should it be 20k?
     
  5. May 14, 2008 #4

    tiny-tim

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    Hi ineedmunchies! :smile:

    You've made it very complicated. :confused:

    ∑(40 - r) from 20 to 40 is simply:

    20 + 19 + … + 1 + 0.

    Just sum that! :smile:
     
  6. May 14, 2008 #5
    I'm pretty sure that should be 19*40 and the 20k should be taken into account too. But as tim said, I've made it too complicated.
     
  7. May 14, 2008 #6
    Tim did make a good observation. Make sure, however, that you don't use 20k. What should it be instead?
     
  8. May 14, 2008 #7
    So can anyone help on how to get the expected value of R?

    I think you multiply the two functions by r and then work out the sum over the ranges. Not entirely sure.

    Oh and does anybody agree with my value of 1/230 for k??


    EDIT: sorry didn't see your post tedjn.

    Should it be 20*19k instead??
     
  9. May 14, 2008 #8
    I think you have the right idea, but how many numbers are between 0 and 19 inclusive?
     
  10. May 14, 2008 #9
    ahhh silly little mistake again. twenty, ok getting there slowly.

    1 = 20*20k + (20*21/2)k = 610k
    k=1/610 hows that look?
     
  11. May 14, 2008 #10
    Yeah, that looks good.
     
  12. May 14, 2008 #11

    tiny-tim

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    Hi ineedmunchies! :smile:

    Yes, that's right!
     
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