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Math Help - Discrete random variable

  1. #1
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    Discrete random variable

    a discrete random variable X has the probability distribution given in the following table. it is given that E(X)=0.95

    x ....... -1 ... 0 ... 1 ...2 ... 3
    P(X=x) ...a...b...0.1...0.3...0.2...


    Find the probability that X is greater than E(X)
    Find values a and b
    Calculate Var(X)
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  2. #2
    Junior Member roy_zhang's Avatar
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    Quote Originally Posted by gracey View Post
    a discrete random variable X has the probability distribution given in the following table. it is given that E(X)=0.95

    x ....... -1 ... 0 ... 1 ...2 ... 3
    P(X=x) ...a...b...0.1...0.3...0.2...


    Find the probability that X is greater than E(X)
    Find values a and b
    Calculate Var(X)

    Realize that finding the probability that X is greater than E(X) is equivalent to find P(X>0.95), from given we know

    P(X>0.95)=P(1)+P(2)+P(3)=0.1+0.3+0.2=0.6

    Now let’s find a and b. By the definition of expected value, we have
    E(X)=\sum\limits_{x=-1}^3 xP(x)=(-1)P(-1)+(0)P(0)+(1)P(1)+(2)P(2)+(3)P(3)
    =-a+0.1+0.6+0.6

    Since E(X)=0.95 we have a=0.35
    Also by probability axiom, we have \sum\limits_{x=-1}^3 P(x)=1, In other words,

    a+b+0.1+0.3+0.2=1\Rightarrow b=0.05

    Lastly, the variance can be calculated by the formula V(X)=E(X^2)-(E(X))^2, note that
    E(X^2)=\sum\limits_{x=-1}^3 x^2P(x)
    =(-1)^2P(-1)+(0)^2P(0)+(1)^2P(1)+(2)^2P(2)+(3)^2P(3)=3.45

    Hence V(X)=E(X^2)-(E(X))^2=3.45-(0.95)^2=2.5475

    Roy
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