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Discrete probability distribution

  • #1

Homework Statement


The problem is as shown in the attatchment.


Homework Equations


The relevant equations are also given in the attatchment.


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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Answers and Replies

  • #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:
  • #3
737
0
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?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
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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:
 
  • #5
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?
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.
 
  • #6
737
0
Tim did make a good observation. Make sure, however, that you don't use 20k. What should it be instead?
 
  • #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??
 
  • #8
737
0
I think you have the right idea, but how many numbers are between 0 and 19 inclusive?
 
  • #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?
 
  • #10
737
0
Yeah, that looks good.
 
  • #11
tiny-tim
Science Advisor
Homework Helper
25,832
249
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.
Hi ineedmunchies! :smile:

Yes, that's right!
 

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