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    hi my friends, i'm really stuck at these 2 questions, which anyone please give me some hints on them?:rolleyes:

    for question c, i am complete clueless...

    for question 8, i know the total area should equal to one. so if i let the area of P(X<80-a) be x, so the area of a and b should be x+2x=1-0.5239
    so x=0.1587
    and i'm stuck from there...

    thanks in advance
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    For c
    P(X=0)=0.2^0 q^{1-0}

    similar for P(X=1)

    P(X=0)+P(X=1)=1.
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    (Original post by BabyMaths)
    For c
    P(X=0)=0.2^0 q^{1-0}

    similar for P(X=1)

    P(X=0)+P(X=1)=1.
    thanks for replying
    but why P(X=0)+P(X=1)=1 ?
    does that mean if i define x=0,1,2 ; i have P(X=0)+P(X=1)+P(X=2)=1 ?
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    (Original post by kingsclub)
    thanks for replying
    but why P(X=0)+P(X=1)=1 ?
    because it should probably say

    ...for x=0, 1 and P(X=x)=0 otherwise.

    X=0 and X=1 are the only possibilities.
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    (Original post by kingsclub)
    hi my friends, i'm really stuck at these 2 questions, which anyone please give me some hints on them?:rolleyes:

    for question c, i am complete clueless...

    for question 8, i know the total area should equal to one. so if i let the area of P(X<80-a) be x, so the area of a and b should be x+2x=1-0.5239
    so x=0.1587
    and i'm stuck from there...

    thanks in advance
    FOr 8
    SO you know
    F(80-a)=0.1587=\Phi (\frac{80-a-80}{13})=\Phi(-\frac{a}{13})=1-\Phi(\frac{a}{13})
    You get a/13 value from the standard distribution table ( \Phi(x)=1-0.1587)

    2F(80-a)=P(X>80+b)=1-P(X<=80+b)=1-F(80+b)
    and similarly to above
 
 
 
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