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Math Help - Variance

  1. #1
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    Variance

    A dealer's profit in units of $5,000 on a new automobile is a random variable X having density function

    f(x,y) = 2(1-x) for -1 <= x <= 1
    a) Find the vaiance in the dealer's profit.
    b) Demonstrate that chebyshev's inequality holds for k = 2 with the density function above.
    c) What is the probability that the profit exceeds $500?

    a) E(2(1-x)) = -4
    \int-1,1 (-2x-1)f(x)dx What is f(x) in this problem?
    b) How can I show that it holds true?
    c) ???
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  2. #2
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    Quote Originally Posted by kenshinofkin View Post
    A dealer's profit in units of $5,000 on a new automobile is a random variable X having density function

    f(x,y) = 2(1-x) for -1 <= x <= 1
    a) Find the vaiance in the dealer's profit.
    b) Demonstrate that chebyshev's inequality holds for k = 2 with the density function above.
    c) What is the probability that the profit exceeds $500?

    a) E(2(1-x)) = -4
    \int-1,1 (-2x-1)f(x)dx What is f(x) in this problem?
    b) How can I show that it holds true?
    c) ???
    f(x)=2(1-x)\ x \in [-1,1]

    is not a density as it integrates up to 4 not 1, are you sure that the range is not [0,1]. Also it seems unlikely that a negative profit would be acceptable on a new car (even less a mean profit that is negative).

    RonL
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  3. #3
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    It could be a typo. So if I go with (0,1) am I doing it right so far. f(x) = x?
    So what about the other parts of the question.
    b) How can I show that it holds true?
    Do I do (1,2) which would show a negitive profit?
    c) ???
    Not sure about this one
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  4. #4
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    Quote Originally Posted by kenshinofkin View Post
    A dealer's profit in units of $5,000 on a new automobile is a random variable X having density function

    f(x,y) = 2(1-x) for 0 <= x <= 1 Mr F says: I don't know why you have f(x , y) ....?
    a) Find the vaiance in the dealer's profit.
    b) Demonstrate that chebyshev's inequality holds for k = 2 with the density function above.
    [snip]

    a) E(2(1-x)) = -4
    \int-1,1 (-2x-1)f(x)dx What is f(x) in this problem?
    b) How can I show that it holds true?
    c) ???
    b) Chebyshev's Inequality: \Pr(|X - \mu| \geq k \sigma) \leq \frac{1}{k^2}.

    So you first need to calculate:

    1. \mu = \int x f(x) \, dx = \int_{0}^{1} x [2(1 - x)] \, dx = 2 \int_{0}^{1} x - x^2 \, dx = \frac{1}{3}.

    2. \sigma^2 = E(X^2) - \mu^2,

    where E(X^2) = \int x^2 f(x) \, dx = \int_{0}^{1} x^2 [2(1 - x)] \, dx = 2 \int_{0}^{1} x^2 - x^3 \, dx = \, ....... (which will answer part a) by the way).

    Therefore \sigma = \frac{1}{3 \sqrt{2}}.


    Substitute \mu = \frac{1}{3}, \sigma = \frac{1}{3 \sqrt{2}} and k = 2 into the left hand side of Chebyshev's Inequality:

    \Pr \left ( \left |X - \frac{1}{3} \right| \geq \frac{\sqrt{2}}{3} \right)


    = \Pr \left( X \geq \frac{\sqrt{2} + 1}{3} \right)


     = \int_{(\sqrt{2} + 1)/3}^{1} 2 (1 - x) \, dx = \, .... \leq \frac{1}{4}


    which demonstrates the inequality.


    Most of this is basic application of routine definitions and formulae.


    NB: I reserve the right for the arithmetic details to contain errors so make sure you carefully check all numerical results given above.
    Last edited by mr fantastic; August 1st 2008 at 06:09 PM.
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  5. #5
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    Quote Originally Posted by kenshinofkin View Post
    A dealer's profit in units of $5,000 on a new automobile is a random variable X having density function

    f(x,y) = 2(1-x) for 0 <= x <= 1
    [snip]
    c) What is the probability that the profit exceeds $500?

    [snip]
    Note that a profit of $500 is equivalent to X = \frac{1}{10} since X is in units of five thousand dollars.

    After this it's all down hill:

    \Pr \left( X > \frac{1}{10} \right) = \int_{1/10}^{1} 2 (1 - x) \, dx = \, ....
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