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Libniz Group Title

Let X be a discrete random variables with PMF given by : P_X(x)= x/15 ,{x=1,2,3,4,5} 0 , otherwise b)let Y=(X-3)^2, find range Y, S_Y , PMF of Y

  • 2 years ago
  • 2 years ago

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  1. bahrom7893 Group Title
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    @amistre64

    • 2 years ago
  2. Libniz Group Title
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    @Zarkon

    • 2 years ago
  3. Zarkon Group Title
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    where are you stuck?

    • 2 years ago
  4. Libniz Group Title
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    P_X(x)= x/15 ,{x=1,2,3,4,5} 0 , otherwise so to find range using Y y= (X-3)^2 ((x/15)-3)^2 y= (1/15 -3)^2 (-44/45)^2 to 0

    • 2 years ago
  5. Libniz Group Title
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    am I right?

    • 2 years ago
  6. Zarkon Group Title
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    no

    • 2 years ago
  7. Zarkon Group Title
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    x takes the values 1,2,3,4,5 y=(x-3)^2 plug those x values into the above equation

    • 2 years ago
  8. Zarkon Group Title
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    for the range(support)

    • 2 years ago
  9. Libniz Group Title
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    oh, it is X, not x

    • 2 years ago
  10. Zarkon Group Title
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    yes...so X takes the values 1,2,3,4,5 Y=(X-3)^2 plug those X values into the above equation :)

    • 2 years ago
  11. Libniz Group Title
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    • 2 years ago
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  12. Zarkon Group Title
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    ok...do what I wrote

    • 2 years ago
  13. Libniz Group Title
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    (1-3)^2,(2-3)^2,(3-3)^2,(4-3)^2,(5-3)^3 4,1,0,1,4

    • 2 years ago
  14. Zarkon Group Title
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    so 0,1,4

    • 2 years ago
  15. Zarkon Group Title
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    that is the range (or support)

    • 2 years ago
  16. Zarkon Group Title
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    now you can find the pmf

    • 2 years ago
  17. Libniz Group Title
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    right?

    • 2 years ago
  18. Libniz Group Title
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    so y=(x-3)^2 do we solve for x?

    • 2 years ago
  19. Zarkon Group Title
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    no

    • 2 years ago
  20. Zarkon Group Title
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    \[P_{Y}(0)=P_{X}(3)\] \[P_{Y}(1)=P_{X}(2)+P_{X}(4)\] \[P_{Y}(4)=P_{X}(1)+P_{X}(5)\]

    • 2 years ago
  21. Libniz Group Title
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    can you explain me how you got that?

    • 2 years ago
  22. Zarkon Group Title
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    \[P_{Y}(1)=P(Y=1)=P(X=2\text{ or }X=4)\] \[=P(X=2)+P(X=4)=P_{X}(2)+P_{X}(4)\]

    • 2 years ago
  23. Zarkon Group Title
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    ok?

    • 2 years ago
  24. Libniz Group Title
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    still figuring out the last post

    • 2 years ago
  25. Zarkon Group Title
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    ok

    • 2 years ago
  26. Libniz Group Title
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    I understand you sum because it is 'or' but why is P(y=1) = P(x=2 or x=4)

    • 2 years ago
  27. Zarkon Group Title
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    \[Y=(X-3)^2\] \[1=(2-3)^2\] \[1=(4-3)^2\] if I tell you Y=1 then either X=2 or X=4

    • 2 years ago
  28. Libniz Group Title
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    got it

    • 2 years ago
  29. Zarkon Group Title
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    good

    • 2 years ago
  30. Libniz Group Title
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    so that's just P(Y=1) do we find Y=1,2,3,4,5?

    • 2 years ago
  31. Zarkon Group Title
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    Y only takes the values 0,1,4

    • 2 years ago
  32. Libniz Group Title
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    got it , thanks; I wish you were my probability professor

    • 2 years ago
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