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Math Help - Geometric distribution and P( E(X) - 2 < E(X) < E(X) + 2)

  1. #1
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    Geometric distribution and P( E(X) - 2 < E(X) < E(X) + 2)

    If we have some random variable X, and it's distributed geometrically.

    X = G(p)

    and

    p = \frac{1}{6}

    To get expected value it's just:

    E(X) = \frac{1}{\frac{1}{6}} = 6

    But how to get this probability:

    P( E(X)-2 < E(X) < E(X)+2 )



    thank you for your help.
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  2. #2
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    Re: Geometric distribution and P( E(X) - 2 < E(X) < E(X) + 2)

    Quote Originally Posted by Nforce View Post
    If we have some random variable X, and it's distributed geometrically.
    X = G(p) and p = \frac{1}{6}
    To get expected value it's just:
    E(X) = \frac{1}{\frac{1}{6}} = 6
    But how to get this probability: P( E(X)-2 < E(X) < E(X)+2 )
    If this is not a trick question, then it is flawed.
    Because for any real number T it is true that \mathcal{P}(T-2<T<T+2)=1.
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    Re: Geometric distribution and P( E(X) - 2 < E(X) < E(X) + 2)

    No, it's not a trick question,

    I am trying to ask what is the probability, that expected value | E(X) - and a result of real experiment | < 2 ?

    we throw a dice and we are looking at just one event: when we throw a 3, for example.

    for this is E(X) = 6. In average we need 6 throws to get one 3 on a dice, then we stop.

    But in a real experiment maybe we had need it 8 throws to get 3 on a dice.

    So what is that probability?
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  4. #4
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    Re: Geometric distribution and P( E(X) - 2 < E(X) < E(X) + 2)

    It reads to me as if you are asking for the P that we need 5,6 or 7 goes (?) However your heading is not correct.

    I think you mean P(E(X)-2<X<E(X)+2)
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