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Libniz

  • 3 years ago

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

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  1. bahrom7893
    • 3 years ago
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    @amistre64

  2. Libniz
    • 3 years ago
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    @Zarkon

  3. Zarkon
    • 3 years ago
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    where are you stuck?

  4. Libniz
    • 3 years ago
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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

  5. Libniz
    • 3 years ago
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    am I right?

  6. Zarkon
    • 3 years ago
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    no

  7. Zarkon
    • 3 years ago
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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

  8. Zarkon
    • 3 years ago
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    for the range(support)

  9. Libniz
    • 3 years ago
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    oh, it is X, not x

  10. Zarkon
    • 3 years ago
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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 :)

  11. Libniz
    • 3 years ago
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  12. Zarkon
    • 3 years ago
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    ok...do what I wrote

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

  14. Zarkon
    • 3 years ago
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    so 0,1,4

  15. Zarkon
    • 3 years ago
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    that is the range (or support)

  16. Zarkon
    • 3 years ago
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    now you can find the pmf

  17. Libniz
    • 3 years ago
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    right?

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

  19. Zarkon
    • 3 years ago
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    no

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

  21. Libniz
    • 3 years ago
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    can you explain me how you got that?

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

  23. Zarkon
    • 3 years ago
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    ok?

  24. Libniz
    • 3 years ago
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    still figuring out the last post

  25. Zarkon
    • 3 years ago
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    ok

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

  27. Zarkon
    • 3 years ago
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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

  28. Libniz
    • 3 years ago
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    got it

  29. Zarkon
    • 3 years ago
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    good

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

  31. Zarkon
    • 3 years ago
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    Y only takes the values 0,1,4

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

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