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Libniz

  • 3 years ago

P(X=-1)= P(x=+1)=1/2P(X=0) a) compute P(X=0) b)compute Standand deviation funcion of x c)draw cumulative distributive function of x d) compute P(x=1| x>=0)

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  1. Libniz
    • 3 years ago
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    P[x=-1]+P(x=1)+P[x=0]=1 P[x=-1]+P[x=-1]=P[x=0] 2P[x=0]=1 P[x=0]=0.5 P[x=1]=P[x=-1]=.25

  2. wio
    • 3 years ago
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    Umm that first line is hard to read.

  3. Libniz
    • 3 years ago
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    b) standard deviations of X E[x]=.25(1)+.25(-1)+0(.5)=0 (E[x])^2=0 E[x^2]=.5 var[x]= .5-0=.5 standard deviation=.5

  4. Libniz
    • 3 years ago
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    c)|dw:1348940632883:dw|

  5. Libniz
    • 3 years ago
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    d) P[x=0]+P[x=1]

  6. Libniz
    • 3 years ago
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    @zarkon

  7. Zarkon
    • 3 years ago
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    your final answer for b is not correct

  8. Libniz
    • 3 years ago
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    oh, sqrt[0.5]

  9. Zarkon
    • 3 years ago
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    correct

  10. Zarkon
    • 3 years ago
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    d)\[P(X=1| X\ge0)=\frac{P(X=1,X\ge 0)}{P(X\ge 0)}\]

  11. Zarkon
    • 3 years ago
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    \[P(X=1| X\ge0)=\frac{P(X=1,X\ge 0)}{P(X\ge 0)}=\frac{P(X=1)}{P(X\ge 0)}\]

  12. Libniz
    • 3 years ago
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    what does the comma mean P(X=1,X≥0)?

  13. Zarkon
    • 3 years ago
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    \[P(X=1 \text{ and } X\ge0)\]

  14. Libniz
    • 3 years ago
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    product of the two then ?

  15. Zarkon
    • 3 years ago
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    no...they are not independent

  16. Zarkon
    • 3 years ago
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    \[P(X=1 \text{ and } X\ge0)=P(X=1)\]

  17. Libniz
    • 3 years ago
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    oh, "and "means it has to satisfy both condition. x=1 is inside x>=0 ; so we choose x=1

  18. Zarkon
    • 3 years ago
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    \[x\ge 0\Rightarrow x\in\{0,1\}\] \[\{1\}\cap\{0,1\}=\{1\}\]

  19. Libniz
    • 3 years ago
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    \[\frac{P(X=1)}{P(X\ge 0)}\] 0.25 ---------- = 1/3 0.75

  20. Zarkon
    • 3 years ago
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    yep

  21. Libniz
    • 3 years ago
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    you are truly a great teacher; thanks sir @Zarkon

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