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    isnt 2) a) 1/3?
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    2(a), P(Xless than 0) = 1/3? find F(0), thus (0+2)/6? is this not rite?
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    (Original post by 183)
    2(a), P(Xless than 0) = 1/3? find F(0), thus (0+2)/6? is this not rite?
    yup thats what i got
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    yh ur right khaggy
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    i definetly got abot 5 mrks wrong
    mayb a few silly mistakes here n there

    wot do u think the grade boundaries wil be for an A
    hopefully aroun 62
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    paper was ok, lost a couple marks due to stupidness tho. hopefuli grade boundaries rnt ludicrously high. 62 would b great
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    (Original post by khaggy)
    yup thats what i got

    Basic rule for probabilities is:

    P(X>=x)=1-P(X<x)

    so

    P(X>=0) = 1-P(X<0)
    P(X>=0)-1 = -P(X<0)
    -(P(X>=0)-1)=P(X<0)

    P(X<0)= -(P(X>=0)-1)

    P(X>=0)=1-P(X<=-1)
    P(X>=0)=1- integrated [(x+2)/6] with limits -1, -2
    P(X>=0)=1-1/12
    P(X>=0)=11/12

    P(X<0)= -(P(X>=0)-1)
    P(X<0)= -(11/12-1)
    P(X<0)= -(11/12-1)
    P(X<0)= 1/12
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    (Original post by Moa)
    Basic rule for probabilities is:

    P(X>=x)=1-P(X<x)

    so

    P(X>=0) = 1-P(X<0)
    P(X>=0)-1 = -P(X<0)
    -(P(X>=0)-1)=P(X<0)

    P(X<0)= -(P(X>=0)-1)

    P(X>=0)=1-P(X<=-1)
    P(X>=0)=1- integrated [(x+2)/6] with limits -1, -2
    P(X>=0)=1-1/12
    P(X>=0)=11/12

    P(X<0)= -(P(X>=0)-1)
    P(X<0)= -(11/12-1)
    P(X<0)= -(11/12-1)
    P(X<0)= 1/12
    P(Xless than or equal to 0) = P(X less than 0)
    as continuous no?

    also as continuous/ rectangular, simply find area less than 0 thus 2*(1/6)
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    Due to it being continuous the basic principle P(X>=x)=1-P(X<x) DOES NOT WORK, as continuous does not need to have an equal sign to be considered as = to.
    Therefore P(x<0) = F(0) = 1/3
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    Has anyone got the paper by any chance?
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    (Original post by balabala4)
    Has anyone got the paper by any chance?
    Edexcel S2 - Tuesday 19 January 2010 Paper: (warning, very poor quality)
    Spoiler:
    Show










    I also posted it on the previous page -.-'
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    (Original post by Moa)
    Edexcel S2 - Tuesday 19 January 2010 Paper: (warning, very poor quality)
    Spoiler:
    Show










    I also posted it on the previous page -.-'

    Im sorry I didnt notice it, but thanks a lot =)... The critical Region i got 17, I must be incorrect as everyone seems to have got 16... but I was always taught P(X<=C2-1)=?... so i gt table value a6 and added 1... bt im nt sure as it states closest to 0.005.
    Please clear my doubt, thanks =)
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    wt did ppl get for the mean for qs 4?
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    (Original post by man u)
    wt did ppl get for the mean for qs 4?
    i got 29/12 = 2.417 others i spoke to got the same
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    (Original post by 183)
    was it not x/3 -1/3
    (Original post by man u)
    I got mostly the same answers but for the cdf i got 1/3x -1/3

    I dont think so, this is what I did:

    Question 4b
    f(x)=3k

    k=1/9

    F(x)=integrated (3k) with limits x,3
    F(x)=k * integrated (3) with limits x,3
    F(x)=1/9 * [3x] limits x,3
    F(x)=1/9 * (3*x - 3*3)
    F(x)=1/9 * (3x -9)
    F(x)=x/3 - 1
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    (Original post by Moa)
    I dont think so, this is what I did:

    Question 4b
    f(x)=3k

    k=1/9

    F(x)=integrated (3k) with limits x,3
    F(x)=k * integrated (3) with limits x,3
    F(x)=1/9 * [3x] limits x,3
    F(x)=1/9 * (3*x - 3*3)
    F(x)=1/9 * (3x -9)
    F(x)=x/3 - 1
    Its cummulative, thus need to add on F(3) which was 2/3
    thus -1/3
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    (Original post by 183)
    Its cummulative, thus need to add on F(3) which was 2/3
    thus -1/3
    This example from the book suggests otherwise:




    I went over the writing to make it easier for you to read, camera in my phone is abysmall Here is the original if you prefer: http://img19.imageshack.us/img19/2121/originallh.jpg
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    (Original post by Moa)
    This example from the book suggests otherwise:




    I went over the writing to make it easier for you to read, camera in my phone is abysmall Here is the original if you prefer: http://img19.imageshack.us/img19/2121/originallh.jpg
    there's only one equation there for the probability density dunction, in the exam, there were two therefore you've got to add on the extra probability from all the ones before.
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    (Original post by Clarity Incognito)
    there's only one equation there for the probability density dunction, in the exam, there were two therefore you've got to add on the extra probability from all the ones before.
    Even if you write separate equations for all ranges?
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    (Original post by Moa)
    Even if you write separate equations for all ranges?
    Essentially, you need to integrate equation one between Xo and its bottom limit. This gives you the cumulative distribution from Xo to the bottom limit.

    For this second equation, you need to integrate equation 2 between Xo and its bottom limit. This gives you the cumulative distribution from Xo to the bottom limit BUT because there is also the extra area of equation 1, you integrate equation 1 between the entirety of its limits. This therefore adds on. In this case, the area from equation 1 is 2/3. Add this to (x/3) - 1 and you get (x/3) - 1/3
 
 
 
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