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Math Help - are my solutions is true???

  1. #1
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    are my solutions is true???

    let X has distribution function given by :

    F(x)=1- e^(-4x) , x>0


    find :

    (1) p(x<=1)

    (2) p(x=4)

    (3) the mean of X

    (4) p (x>=25)

    (5) the second moment about the mean

    my solutions :

    (1) 1-(e^-4)

    (2) zero since the distribution is continuous

    (3)mu=(1\16)

    (4)p (x>=25)= 1 - p (x<=25)

    = 1- { 1 - e^(-4*25) }

    (5) moment generating function is :

    ( 1 - (t\4) )^-1 we differentiate it twice and substitute t=0





    ops: ops: :P
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  2. #2
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    Quote Originally Posted by flower3 View Post
    let X has distribution function given by :

    F(x)=1- e^(-4x) , x>0


    find :

    (1) p(x<=1)

    (2) p(x=4)

    (3) the mean of X

    (4) p (x>=25)

    (5) the second moment about the mean

    my solutions :

    (1) 1-(e^-4) Mr F says: Correct.

    (2) zero since the distribution is continuous Mr F says: Correct.

    (3)mu=(1\16) Mr F says: I get 1/4. Note that the pdf is {\color{red}f(x) = 4 e^{-4x}}, x > 0, and zero otherwise. Note also the result you get when you differentiate the moment generating function (below) and substitute t = 0 ......

    (4)p (x>=25)= 1 - p (x<=25)

    = 1- { 1 - e^(-4*25) } Mr F says: Correct. And I'm sure you can simplify it.

    (5) moment generating function is :

    ( 1 - (t\4) )^-1 Mr F says: Correct.

    we differentiate it twice and substitute t=0 Mr F says: This gives you the second moment about the origin. However, the mean is not equal to zero and so this will not equal the second moment about the mean.
    [snip]
    Since the mean is equal to 1/4, the second moment about the mean is E\left[ \left(X - \frac{1}{4}\right)^2\right].

    In fact, the second moment about the mean is equal to the variance of X. And you know that Var(X) = E(X^2) - \mu^2 = E(X^2) - \frac{1}{16} where E(X^2) is the second moment about the origin ......
    Last edited by mr fantastic; August 6th 2008 at 10:27 PM.
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  3. #3
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    thanks for you
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