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Thread: Calculating Expected Value of a Continuous Random Variable

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    Calculating Expected Value of a Continuous Random Variable




    Hell all,

    I have this distribution function of a random variable X.

    I wish to find E(X).

    I have used derivatives to get the density function, compared it to 1, and found that f(t) = (4/5)t+(3/5). I then used integral of tf(t) over the range of 0 to 1, and got 0.56667. According to the answer I got, it's incorrect (maybe the answer is incorrect, not me?). Can you please assist? Thank you !
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    Re: Calculating Expected Value of a Continuous Random Variable


    Assuming C is a constant, I believe that the expected value of the prob. density function would be: 4/15 + C/2.

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