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The Official Edexcel S2 Revision Thread! (22/06/05 PM) watch

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    would the answer to that soloman paper question be 4/9 or 5/9?

    I cant figure out whether u include the probabilities of 3 and 7 (i.e. 3+4+5+6+7) or if u just do 7-3 * 1/9
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    for the soloman paper are the first answers
    mean=13/2=6.5

    P(x>=8.6)
    =(11-8.6)xh
    where h=1/9
    P=0.267

    c)p(Ix-5I<2)=P(3<x<7)
    =1/9 x 4 =4/9

    ?and can someone please give me a quick explanation of how to get the var equal to what they give??

    all i have in notes for calculating var is

    Var(x)=E(x^2)-(mean^2)
    where e(x^2) =integ b->a (x^2)/b-a

    I have tried this and end up with lots of a's and b's that dont cancel down to what is given, please someone say there is a similar way of figuring out variance for continuous-uniform-dist.
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    Can anyone do this one? - it's a "multi pdf" that turns into a "multi cdf" ...maybe...
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    (Original post by Swordsikan)
    Can anyone do this one? - it's a "multi pdf" that turns into a "multi cdf" ...maybe...
    argh even the first part keeps going wrong, i found k to br -1/6 grrrr
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    ok, I sort of worked it out...umm it will probably take all day to type it up, and my scanner is down so I'll just try and explain.
    I'll leave out the "show" bit because you can do it with the CDF, also if you draw a graph you can show that it is symmetrical and so k must be 1/3...

    This is fairly new to me, but what you can do work out C is integrate the first line as follows
    ..x
    (1/9) |t -2 dt => (1/9) (((X^2)/2)-(-2)
    ..2

    => (1/18)((X^2) -4x +4)

    for the next line you do the same, but you integrate between 5 and x. and you add 0.5 which is the value you get out when you put 5 into the above integral...This is pretty confusing as it is...but showing it here makes it 100 times worse. Hope someone understands it.
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    (Original post by Swordsikan)
    ok, I sort of worked it out...umm it will probably take all day to type it up, and my scanner is down so I'll just try and explain.
    I'll leave out the "show" bit because you can do it with the CDF, also if you draw a graph you can show that it is symmetrical and so k must be 3...

    afk..I'll finish later
    Yeah, i got it by ignoring the show part first. But is there a way of doing it without having the c.d.f? If there is plz explain.
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    (Original post by Swordsikan)
    Can anyone do this one? - it's a "multi pdf" that turns into a "multi cdf" ...maybe...
    where did you get this question from, i cant even do the first bit.
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    (Original post by Swordsikan)
    Can anyone do this one? - it's a "multi pdf" that turns into a "multi cdf" ...maybe...
    integrate with respect to x n add..with limits 5,2 n 8,5 respectively
    k/3 { ∫ (x - 2) dx + ∫ (8-x) dx } = 1
    [x²/2 - 2x] + [8x - x²/2] = 3/k
    (4.5 + 4.5) k = 3
    k= 3/9 = 1/3

    second bit:

    F(x) = 1/9 (x²/2 - 2x + c) 2<x<5
    when x=2 F(x)=0, giving c = 2

    F(x) = 1/9 (8x - x²/2 + k) 5≤x<8
    when x=8 F(x)=1, giving k = -23

    [u shud check by seeing its a 'smooth' transition from one bit of the cdf to the other...probabilities shud come out same for upper boundary of the previous bit n the lower one of next -- does that even make sense? :rolleyes: ]

    anyway so overall
    F(x) = 0 x≤2
    --------1/9 (x²/2 - 2x + 2) 2<x<5
    --------1/9 (8x - x²/2 - 23) 5≤x<8
    --------1 x≥8
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    The question is from the S2 revision guide test paper at the back (edexcel heineman) I cut it down a bit becuase it takes ages to write out. The second part of the question is actually a "show that CDF looks like this" question..but I figured it may not be like that in the actual exam (not to mention it would take me ages to type up).
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    o n the last bit,

    just plug in 4 and 7 into the cdf n subtract answers
    so its
    [1/9 (8*7 - 7²/2 - 23)] - [1/9 (8*4 - 4²/2 - 23)] = 0.8333
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    (Original post by hajira)
    o n the last bit,

    just plug in 4 and 7 into the cdf n subtract answers
    so its
    [1/9 (8*7 - 7²/2 - 23)] - [1/9 (8*4 - 4²/2 - 23)] = 0.8333
    genius!
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    (Original post by hajira)
    o n the last bit,

    just plug in 4 and 7 into the cdf n subtract answers
    so its
    [1/9 (8*7 - 7²/2 - 23)] - [1/9 (8*4 - 4²/2 - 23)] = 0.8333

    I think the last answer is supposed to be 13/18, but the other bit seems right.

    Here is a bit at the end of that question:
    Find the lower quartile of the distribution
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    I've just had a thought, with multi pdf thingies, do they ask, and furthermore, what does one do if they ask for the mean and the variance? Is it just top of top range minus bottom of bottom range with definite integrals?
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    (Original post by Swordsikan)
    I think the last answer is supposed to be 13/18, but the other bit seems right.

    Here is a bit at the end of that question:
    Find the lower quartile of the distribution
    o yeah..my bad
    4 is in the other cdf
    so it shud b
    [1/9 (8*7 - 7²/2 - 23)] - [1/9 (4²/2 - 2*4 + 2)] = 6.5/9 = 13/18

    sorry!

    for the last bit..ud just find the value of x for which F(x)=0.25
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    Ok I think what you did wrong for the last section P(4<= x <= 7) that is, is that you need to plug the 4 into the top CDF line, and the 7 into the lower one.
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    (Original post by Swordsikan)
    Ok I think what you did wrong for the last section P(4<= x <= 7) that is, is that you need to plug the 4 into the top CDF line, and the 7 into the lower one.
    yeah iv corrected it now!
    edit: woops ur reply is at the same time

    anyway as lower quartile is more likely to be in the top range, using that to get F(x)=0.25 shud get x value for lower quartile

    im too tired to type it up!
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    can i just ask..... critical regions! help!
    if they ask for the closest to 2.5% say you find the two that are closest to 0.025 and 0.975 (i.e. on eiether side of those values) as the critical region.
    but if they say find a critical region to 5% sig do you find above 0.025 and below 0.975??? or not... i am confused with this because every question i do one way they want the other!!
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    ok thank you, for your help..although how would you go about knowing that the lower quartile is in which section of the CDF?
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    (Original post by green_eyes)
    can i just ask..... critical regions! help!
    if they ask for the closest to 2.5% say you find the two that are closest to 0.025 and 0.975 (i.e. on eiether side of those values) as the critical region.
    but if they say find a critical region to 5% sig do you find above 0.025 and below 0.975??? or not... i am confused with this because every question i do one way they want the other!!
    it depends on whether the test is one-tailed or two-tailed..
    if its two n they ask for 5%...then u will look at closest to 0.025 n 0.975 at either end
    however if its one-tailed n 5% significance is asked for, u'll look at 0.95 or 0.05, depending on whether its > or <
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    (Original post by Swordsikan)
    ok thank you, for your help..although how would you go about knowing that the lower quartile is in which section of the CDF?
    well that wud usually be clear from a graph if u'v drawn one
    or if uv calculated the mean..the lower quartile is likely to be below it n the higher above it
    however where u hav neither like in this case, its more of an 'intelligent guess' i suppose, by looking at the range of values of x
    at least thats wat i was told by my teacher!
 
 
 
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