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Math Help - distribution function with uniformity

  1. #1
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    distribution function with uniformity

    I'm having trouble solving the following problem correctly:

    The waiting time Y until delivery of a new component for an industrial operation is uniformly distributed over the interval 1 to 5 days. The cost of this delay is given by U = 2Y^2+3. Find the probability density function for U.

    so far I have:

    Y= \pm \sqrt{\frac{u-3}{2}}

    and based on an example from the book I have:

    f_U(u) = \left\{ \begin{array}{rcl}<br />
\frac{1}{2\sqrt{u}} \left[f_Y\left(\sqrt{u}\right) + f_Y(y)(-\sqrt{u})\right]  & \mbox{for} & u>0 \\ <br />
0 & \mbox{if} & \mbox{other} <br />
\end{array}\right.

    then combining for the top part I would get:

    =\frac{1}{2\sqrt{\frac{u-3}{2}}} \left[\frac{1}{4}\left(\sqrt{\frac{u-3}{2}}\right) + \frac{1}{4}\left(-\sqrt{\frac{u-3}{2}}\right)\right]

    =\frac{1}{8\sqrt{\frac{u-3}{2}}}

    and based on the solution in the back of the book it's:

    =\frac{1}{{\color{red}16}\sqrt{\frac{u-3}{2}}}

    if anyone can help me it it would be greatly appreciated
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  2. #2
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    Quote Originally Posted by lllll View Post
    I'm having trouble solving the following problem correctly:

    The waiting time Y until delivery of a new component for an industrial operation is uniformly distributed over the interval 1 to 5 days. The cost of this delay is given by U = 2Y^2+3. Find the probability density function for U.

    so far I have:

    Y= \pm \sqrt{\frac{u-3}{2}}

    and based on an example from the book I have:

    f_U(u) = \left\{ \begin{array}{rcl}<br />
\frac{1}{2\sqrt{u}} \left[f_Y\left(\sqrt{u}\right) + f_Y(y)(-\sqrt{u})\right] & \mbox{for} & u>0 \\ <br />
0 & \mbox{if} & \mbox{other} <br />
\end{array}\right.

    then combining for the top part I would get:

    =\frac{1}{2\sqrt{\frac{u-3}{2}}} \left[\frac{1}{4}\left(\sqrt{\frac{u-3}{2}}\right) + \frac{1}{4}\left(-\sqrt{\frac{u-3}{2}}\right)\right]

    =\frac{1}{8\sqrt{\frac{u-3}{2}}}

    and based on the solution in the back of the book it's:

    =\frac{1}{{\color{red}16}\sqrt{\frac{u-3}{2}}}

    if anyone can help me it it would be greatly appreciated
    F(u) = \Pr(U \leq u)

     = \Pr(2Y^2 + 3 \leq u) = \Pr \left( Y^2 \leq \frac{u-3}{2} \right)  = \Pr \left( - \sqrt{\frac{u-3}{2}} \leq Y \leq \sqrt{\frac{u-3}{2}} \right)


    = \int_{1}^{\sqrt{\frac{u-3}{2}}} \frac{1}{4} \, dy


    = \frac{1}{4} \sqrt{\frac{u-3}{2}} - \frac{1}{4}.


    Therefore f(u) = \frac{dF}{du} = \frac{1}{16\sqrt{\frac{u-3}{2}}} for 5 \leq u \leq 53 and zero otherwise.


    Note: You can see your answer is wrong because if you integrate it between u = 5 and u = 53 you get 2 rather than 1.
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