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Math Help - Probability Question regarding determining the constant in the pdf

  1. #1
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    Probability Question regarding determining the constant in the pdf

    Question : A random variable X has a pdf

    f(x) = \begin{cases}<br />
 Ax , & \mbox{0 } \le \mbox{ x} < \mbox{ 5} \\ <br />
A(10 -x) , & \mbox{5} \le \mbox{ x} < \mbox{10} \\ <br />
0, & \mbox{otherwise }<br />
\end{cases}

    i) Determine A

    ii) Find P(2.5 \le X \le 7.5)
    Last edited by zorro; December 13th 2009 at 07:49 PM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by zorro View Post
    Question : A random variable X has a pdf

    f(x) = \begin{cases}<br />
 Ax , & \mbox{0 } \le \mbox{ x < 5} \\ <br />
A(10 -x) , & \mbox{5} \le \mbox{ x <10} \\ <br />
0, & \mbox{otherwise }<br />
\end{cases}

    i) Determine A

    ii) Find P(2.5 \le X \le 7.5)
    What attempt have you made to this problem? First note that if f is a pdf, then \int_{-\infty}^{\infty}f\!\left(x\right)\,dx=1...

    To answer the second part, use the definition of the pdf.
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  3. #3
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    I am having problem with the first part

    Quote Originally Posted by Chris L T521 View Post
    What attempt have you made to this problem? First note that if f is a pdf, then \int_{-\infty}^{\infty}f\!\left(x\right)\,dx=1...

    To answer the second part, use the definition of the pdf.
    i am having with determining the value of A hw should i go about it
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by zorro View Post
    i am having with determining the value of A hw should i go about it
    Integrate each part of the piecewise function over its domain, and then set your answer equal to 1. Then solve the equation you created for A.
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    Is this correct?

    Quote Originally Posted by Chris L T521 View Post
    Integrate each part of the piecewise function over its domain, and then set your answer equal to 1. Then solve the equation you created for A.

    Is this correct?

    f(0 < x < 5) \ = \ \int_{0}^{5} Ax \ = \ \frac{25}{2}A


    f(5 < x < 10) \ = \ \int_{5}^{10} A(10 - x) \ = \ \frac{25}{2} A

    \frac{25}{2}A \ = \ 1

    \therefore \ \ A = \frac{2}{25}
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by zorro View Post
    Is this correct?

    f(0 < x < 5) \ = \ \int_{0}^{5} Ax \ = \ \frac{25}{2}A


    f(5 < x < 10) \ = \ \int_{5}^{10} A(10 - x) \ = \ \frac{25}{2} A

    \frac{25}{2}A \ = \ 1

    \therefore \ \ A = \frac{2}{25}
    Close... You need to add both of the values together and then set equal to 1. So A=\ldots
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    Is this correct?

    Quote Originally Posted by Chris L T521 View Post
    Close... You need to add both of the values together and then set equal to 1. So A=\ldots


    \frac{25}{2} A +  \frac{25}{2} A \ = \ 1

    \frac{50}{2} A \ = \ 1

    \frac{50}{2} A \ = \ 1

    25 A = \ 1

    \therefore \ A \ = \ \frac{1}{25}


    Now for

    P(2.5 \le X \le 7.5) \ = \ \int_{2.5}^{5} Ax \ dx+ \int_{5}^{7.5} A(10-x)dx

    Is this correct???
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by zorro View Post
    \frac{25}{2} A +  \frac{25}{2} A \ = \ 1

    \frac{50}{2} A \ = \ 1

    \frac{50}{2} A \ = \ 1

    25 A = \ 1

    \therefore \ A \ = \ \frac{1}{25}


    Now for

    P(2.5 \le X \le 7.5) \ = \ \int_{2.5}^{5} Ax \ dx+ \int_{5}^{7.5} A(10-x)dx

    Is this correct???
    Correct! But note that you know what A is...so you should evaluate \int_{2.5}^{5} \tfrac{1}{25}x\,dx+ \int_{5}^{7.5} \tfrac{1}{25}(10-x)\,dx.
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  9. #9
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    Thanks Chris L T521

    Quote Originally Posted by Chris L T521 View Post
    Correct! But note that you know what A is...so you should evaluate \int_{2.5}^{5} \tfrac{1}{25}x\,dx+ \int_{5}^{7.5} \tfrac{1}{25}(10-x)\,dx.

    thanks Chris L T521 for helping me so much
    Cheers mite

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