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Math Help - 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:
    f(x)\ =\ \frac{1}{2}\ for\ x\ between\ 0\ and\ 1
    \          \frac{3\ -\ x}{4}\ for\ x\ between\ 1\ and\ 3

    \          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 \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
    \int{\frac{3\ -\ x}{4}}\ from\ 1\ to\ x, which becomes:
    [\frac{1}{4}(3x\ -\ x^2)] with limits 1 and x.

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

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

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

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

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

    However, the actual answer is \frac{119}{128}, and the cumulative distribution function F(x) is 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:
    f(x)\ =\ \frac{1}{2}\ for\ x\ between\ 0\ and\ 1
    \          \frac{3\ -\ x}{4}\ for\ x\ between\ 1\ and\ 3

    \          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 \frac{13}{12}.
    d) Find P(X < 3u - n).
    3u-n=2.25

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

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

    \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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