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S2 - Help Needed on Multi-Parts CDF Question watch

    • Thread Starter

    Im currently doing the last question on the S2 June 2002 paper
    Q7 (b) (i)

    It is a 3-part PDF and they want me to find F(x) for the 3rd part.

    I know I have to intergrate it with the limits 'x0' and the lower boundry of the range to get F(x), but you also have to add something to your intergrated answer at the end...like a follow thru from the previous parts of the PDF.

    Do you add the value of F(x) of the 2nd part, or do you add both values of F(x) the 1st AND 2nd parts?

    I know it sounds confusing but if someone understands what im going on about then i would be greatful!

    I recognize what your saying I think, so I think I know the question

    Off the top of my head, integrate the middle bit, and use the integration of the upper bit to find the cdf of the value 1...then, with the middle intergration, say that (blah) + C = whatever the first integration gave you.....i think this is it :p:
    • Thread Starter

    Sounds a bit confusing...

    Put it this way:

    When theres a PDF with 2 parts (each with a different range), when asked to find the CDF you:

    1) intergrate the first part of the PDF which gives you F(x) of the first bit.
    2) then intergrate the second bit of the PDF, (then putting the upper boundry of the first range into the first F(x) - then add this to the intergration of the second part)

    This is easy, but what im confused about is whens there is 3 parts.
    Once intergrated the 3rd part of the PDF, do you add to it 2 values (from F(x) of 1st and 2nd part) OR add 1 value (F(x) from ONLY the 2nd part)

    Ooops, wrong paper, will attempt to answer the question spontaneously...


    For the bottom range, it is just a matter of integrating, as it starts at 0 and there is no arbitrary constant of integration...therefore..

    = x²/15

    For the top range, when x=10, F(x) must =1, as 10 is the top of the range, and it encompasses all probabilities....with that in mind...

    First stage : = 4x/9 - x²/45 + C

    now, F(10) = 1

    => 40/9 - 100/45 + C = 1

    => 20/9 +C =1

    => C = -11/9

    therefore, the end equation = 4x/9 - x²/45 - 11/9
    • Thread Starter

    thats great, thanks
    i was just attemtping it a different way which my teacher told me. im still not sure about the method my teacher told me, but your way seems to do the job

    You listened to your statistics teacher?? ...if I did that with my stats teacher, it would be a death wish :p:
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