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    The random variable f(x) is given by f(x) = kx(16-x^2) for 0<_x<_4


    a) find the value of k.

    b) find F(X)
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    (Original post by Tempa)
    The random variable f(x) is given by f(x) = kx(16-x^2) for 0<_x<_4


    a) find the value of k.

    b) find F(X)
    Show some working, this is a pretty standard question in terms of method.

    Also, for (b), do you mean find E(X)?
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    (Original post by Tempa)
    The random variable f(x) is given by f(x) = kx(16-x^2) for 0<_x<_4


    a) find the value of k.

    b) find F(X)
    The area under this function is the probability.
    The total area will be equal to one, as one is the total possible amount.
    Therefore, the integral of the function between 0 and 4 will be equal to one.
    You can now find k.
    ???
    Profit.

    _Kar.
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    (Original post by Farhan.Hanif93)
    Show some working, this is a pretty standard question in terms of method.

    Also, for (b), do you mean find E(X)?
    No it is F(x)

    The method is to integrate f(x) and put in the 4 and 0 as x. Then take it away, but I keep getting the wrong answer.

    16kx - kx^2 = 1

    integrated = 8kx^2 - kx^3/3 =1

    8k x 4^2 - k x 4^3 / 3 - 8k x 0 - k x 0/3 =1

    128k - 64/3k = 1

    And then you get some dodgy answer for k.
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    (Original post by Tempa)
    No it is F(x)

    The method is to integrate f(x) and put in the 4 and 0 as x. Then take it away, but I keep getting the wrong answer.

    16kx - kx^2 = 1

    integrated = 8kx^2 - kx^3/3 =1

    8k x 4^2 - k x 4^3 / 3 - 8k x 0 - k x 0/3 =1

    128k - 64/3k = 1

    And then you get some dodgy answer for k.
    It's because the integration is

    \displaystyle\int 16kx-kx^3dx

    I think you made an error in the first line

    _Kar.
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    (Original post by Kareir)
    It's because the integration is

    \displaystyle\int 16kx-kx^3dx

    I think you made an error in the first line

    _Kar.
    f(x) = kx(16-x^2) for 0<_x<_4

    What did u get when u multiplied out the brackets ?

    I got 16kx - kx^2
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    (Original post by Tempa)
    f(x) = kx(16-x^2) for 0<_x<_4

    What did u get when u multiplied out the brackets ?

    I got 16kx - kx^2
     kx(16-x^2) =



kx\times16 - kx\times x^2= 



16kx - kx^3

    _Kar.
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    (Original post by Kareir)
     kx(16-x^2) =



kx\times16 - kx\times x^2= 



16kx - kx^3

    _Kar.
    **** sakes I made the silliest mistake, which wasted 30mins of my life.

    But thanks m8, for helping me out.
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    (Original post by Tempa)
    **** sakes I made the silliest mistake, which wasted 30mins of my life.

    But thanks m8, for helping me out.
    De Nada.

    _Kar.
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    (Original post by Tempa)
    The random variable f(x) is given by f(x) = kx(16-x^2) for 0<_x<_4


    a) find the value of k.

    b) find F(X)
    Take the k outside the integral then integrate, put this equal to 1, which, to be fair quite often gives a vile answer then to get F(x) you just integrate between x and 0.
 
 
 
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