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Thread: Cumulative distribution function integration

  1. #1
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    Cumulative distribution function integration

    The Question:
    The continuous random variable X has probability density function given by:
    $\displaystyle f(x)\ =\ \frac{1}{2}\ for\ x\ between\ 0\ and\ 1$
    $\displaystyle \ \frac{3\ -\ x}{4}\ for\ x\ between\ 1\ and\ 3$

    $\displaystyle \ 0\ otherwise$

    a) Sketch the graph of f.
    b) Explain why the value of n, the median of X, is 1.
    c) Show that the value of u, the mean of X, is $\displaystyle \frac{13}{12}$.
    d) Find P(X < 3u - n).

    I've done a), b), and c), but am stuck on d).

    My Attempt:
    F(x) (the cumulative distribution function for X) is
    $\displaystyle \int{\frac{3\ -\ x}{4}}\ from\ 1\ to\ x$, which becomes:
    $\displaystyle [\frac{1}{4}(3x\ -\ x^2)]$ with limits 1 and x.

    Substituting:
    $\displaystyle \frac{3x\ -\ x^2}{4}\ -\ \frac{5}{8}$.

    Now 3u - n = $\displaystyle \frac{9}{4}$.

    Hence $\displaystyle F(\frac{9}{4})$ =:

    $\displaystyle \frac{3(\frac{9}{4})\ -\ \(\frac{9}{4})^2}{4}\ - \frac{5}{8}$.

    This works out to $\displaystyle \frac{55}{128}$.

    However, the actual answer is $\displaystyle \frac{119}{128}$, and the cumulative distribution function F(x) is $\displaystyle 1\ -\ \frac{1}{8}(3\ -\ x)^2$.

    Some advice on how to proceed would be appreciated.

    isx99
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by isx99 View Post
    The Question:
    The continuous random variable X has probability density function given by:
    $\displaystyle f(x)\ =\ \frac{1}{2}\ for\ x\ between\ 0\ and\ 1$
    $\displaystyle \ \frac{3\ -\ x}{4}\ for\ x\ between\ 1\ and\ 3$

    $\displaystyle \ 0\ otherwise$

    a) Sketch the graph of f.
    b) Explain why the value of n, the median of X, is 1.
    c) Show that the value of u, the mean of X, is $\displaystyle \frac{13}{12}$.
    d) Find P(X < 3u - n).
    $\displaystyle 3u-n=2.25$

    $\displaystyle P(X< 3u-n)=P(X<2.25)=\int_0^{2.25} f(x) \; dx$

    Now as $\displaystyle 1<2.25<3$ the integral may be written:

    $\displaystyle \begin{array}{lll}P(X<2.25)&=&{\displaystyle \int_0^{1} f(x) \; dx+\int_1^{2.25} f(x) \; dx} \\ \\ &=&{\displaystyle \frac{1}{2}+\int_1^{2.25}\frac{3-x}{4}\;dx} \end{array}$

    CB
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  3. #3
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    Thank you for your reply

    Thank you for your reply. This is a much simpler method than the one I was using, and it is also much easier to understand.

    isx99
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