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Math Help - Continuous Random Variable with Distribution!

  1. #1
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    Unhappy Continuous Random Variable with Distribution!

    P(x)={k(1-x^3) 0< or = x < or = 1
    0 otherwise

    Evaluate k
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Pringles View Post
    P(x)={k(1-x^3) 0< or = x < or = 1
    0 otherwise

    Evaluate k
    For this to be a pdf, \int_{-\infty}^{\infty}P\left(x\right)\,dx=1

    Thus, solve the equation \int_{-\infty}^{\infty} k\left(1-x^3\right)\,dx=1\implies \int_0^1 k\left(1-x^3\right)\,dx=1 for k.

    Can you take it from here?
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  3. #3
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    I'll try

    Quote Originally Posted by Chris L T521 View Post
    For this to be a pdf, \int_{-\infty}^{\infty}P\left(x\right)\,dx=1

    Thus, solve the equation \int_{-\infty}^{\infty} k\left(1-x^3\right)\,dx=1\implies \int_0^1 k\left(1-x^3\right)\,dx=1 for k.

    Can you take it from here?
    I'll try.Kindly check on me again.thanks
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  4. #4
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    Smile Asswer?

    [x-1/4X^4]0 1

    [1-1/4]-[0]

    =3/4

    I hope it is right.WHAT ABOUT THE PRESENTATION

    thanks
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Pringles View Post
    [x-1/4X^4]0 1

    [1-1/4]-[0]

    =3/4

    I hope it is right.WHAT ABOUT THE PRESENTATION

    thanks
    Recall that k{\color{red}\int_0^1 \left(1-x^3\right)\,dx}=1

    You got the correct value for the part in red. Now what's k?
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  6. #6
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    K=4/3 ?

    so K=4/3
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Pringles View Post
    so K=4/3
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