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Thread: Expected Value

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    Expected Value

    In a company, we manufacture metal poles. Suppose the length of a pole is a random variable X, with mean $\displaystyle m_{X}$ and probability density function $\displaystyle f_{X}(x)$. Poles are cut to obtain an exact length L. If the initial length of the pole is less than L, the entire pole is lost. If it is greater than L, the pole will be cut down to L, and the section left over is lost.

    We are interested in random variable Y, defined as the length of each piece lost.

    (i) Sketch the graph of y (values of Y) as a function of x (values of X), and derive $\displaystyle m_{Y} = \mathbb{E}(Y)$ as a function of $\displaystyle f_{X}(x)$ and $\displaystyle m_{X}$.

    (ii) Suppose X follows a normal distribution with mean $\displaystyle m_{X}$ and variance $\displaystyle \sigma^{2}_{X}$ . Show there exists a value $\displaystyle m_{0}$ of $\displaystyle m_{X}$ which makes $\displaystyle m_{Y}$ minimum.

    (iii) Let L = 2m, $\displaystyle \sigma$ = 0.02m. What is the value of $\displaystyle m_{X}$ which minimizes the amount of lost material.

    Thanks a lot for your help!

    Original Post:

    http://www.mathhelpforum.com/math-he...obability.html

    I have the same question here, and I am stuck at part ii, I have done the first part and get the results

    m(Y) = m(X) - L + LF(x)

    to find the min. m(Y) I think I have to differentiate both sides...but how can I differentiate "LF(x)"...? And I couldnt find any solution in the original post.

    Can anyone help plz?
    Last edited by sweetadam; May 6th 2010 at 07:07 AM.
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