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Math Help - Discrete Random Variables

  1. #1
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    Discrete Random Variables

    Given E[X]=0.95. Find the probability that X is greater than E[X].

    x -1 0 1 2 3
    P(X=x) 0.35 0.05 0.1 0.3 0.2

    Unsure what's asking me there.

    thanks for the help!
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  2. #2
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    Maybe P(X>0.95) = 1 - P(less\ then\ 0.95) = 1 - P(-1) + P(0) = 1 - 0.4 = 0.6 ?
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  3. #3
    Super Member Failure's Avatar
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    Quote Originally Posted by BabyMilo View Post
    Given E[X]{\color{red}=}0.95. Find the probability that X is greater than E[X].

    x -1 0 1 2 3
    P(X=x) 0.35 0.05 0.1 0.3 0.2

    Unsure what's asking me there.

    thanks for the help!
    Well, exactly what they say, I suppose, i.e. they want to know

    P(X>E[X]){\color{red}=}P(X>0.95)=P(X \in \{1,2,3\})
    =P(X=1)+P(X=2)+P(X=3)=0.1+0.3+0.2=0.6.
    BTW. It's a bit strange that they tell you what E[X] is, because from the given values and probabilities of X one can easily determine the expected value.
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  4. #4
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    Quote Originally Posted by BabyMilo View Post
    Given E[X]=0.95. Find the probability that X is greater than E[X].

    x -1 0 1 2 3
    P(X=x) 0.35 0.05 0.1 0.3 0.2

    Unsure what's asking me there.

    thanks for the help!
    Hi BabyMilo,

    E[X]=-1(0.35)+0(0.05)+1(0.1)+2(0.3)+3(0.2)=0.95

    All the probabilities sum to 1.

    The probability that X>E[X] is the probability of x being 1, 2 or 3,
    as those values of x are >0.95

    which is 0.1+0.3+0.2
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  5. #5
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    Quote Originally Posted by Failure View Post
    BTW. It's a bit strange that they tell you what E[X] is, because from the given values and probabilities of X one can easily determine the expected value.
    Yes, that is unusual
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