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Math Help - Continuous Probability functions

  1. #1
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    Continuous Probability functions

    http://myasdcommunity.com/continuous.jpeg
    (I keep getting not a valid image file, sorry, so couldn't display :S...)

    I've managed the first two parts of the question. How ever i was a bit confused as for how the second question the solutions managed to obtain:

    1- \frac {1}{x^2}


    For the integral of the function.

    Thus for the third part of the question, i am confused:
    It would make sense to have to work out P(X >= 4) which would be
    1 - P(X <= 3)

    However they sub in 4 into an equation and get:
    1 - FT (4)  = 1 - (1 - (1 -  \frac{1}{4^2} )= 1/16

    And i am very confused as to where this comes from :S



    Thank-you.
    Last edited by AshleyT; May 12th 2011 at 04:36 PM.
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  2. #2
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    Quote Originally Posted by AshleyT View Post
    http://myasdcommunity.com/continuous.jpeg
    (I keep getting not a valid image file, sorry, so couldn't display :S...)

    I've managed the first two parts of the question. How ever i was a bit confused as for how the second question the solutions managed to obtain:

    1- \frac {1}{x^2}


    For the integral of the function.

    Thus for the third part of the question, i am confused:
    It would make sense to have to work out P(X >= 4) which would be
    1 - P(X <= 3)

    However they sub in 4 into an equation and get:
    1 - FT (4) = 1 - (1 - (1 - \frac{1}{4^2} )= 1/16

    And i am very confused as to where this comes from :S



    Thank-you.
    The solutions are using the cdf of X:


    cdf = F(x) = \int_1^x \frac{d}{t^3} \, dt = \frac{d}{2} \left(1 - \frac{1}{x^2}\right)

    Since F(+\infty) = 1 it's clear that d = 2.


    So F(x) = 1 - \frac{1}{x^2}.


    Then you require \Pr(X \geq 4) = 1 - \Pr(X < 4) = 1 - F(4).


    I suggest you review the definition, calculation and application of the cumulative distribution function cdf.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    The solutions are using the cdf of X:


    cdf = F(x) = \int_1^x \frac{d}{t^3} \, dt = \frac{d}{2} \left(1 - \frac{1}{x^2}\right)

    Since F(+\infty) = 1 it's clear that d = 2.


    So F(x) = 1 - \frac{1}{x^2}.


    Then you require \Pr(X \geq 4) = 1 - \Pr(X < 4) = 1 - F(4).


    I suggest you review the definition, calculation and application of the cumulative distribution function cdf.
    Thank-you.
    I've been reading notes on everything, however because i have ADD, i have to learn by example and have been struggling to find any except for a few questions which had poorly written solutions with no explanations. When i was taught it 3 years ago, i was taught to do it differently which is why i was getting confused, assuming they were doing things the same way.

    I must have made an error however as i was not achieving the same answer at 1/16 with my method.

    Completely understand now, thankyou .
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