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    Hey, I thought it was quite an easy paper compared to m1 exam on wedn. ))

    I didn't have time to write down all the answers... the ones i have are:

    q1, sd=23.31
    iqr=25.5 (upper was around 20 and lower was negative)

    q2, part iv) 0.064


    q4, 2 tail test, 2.5% on each tail. for critical regions i got x>=7


    How did it go for you guys?
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    Hey,

    I got exactly the same critical region as u as well as IQR and s.d.. I found that exam quite difficult overall.
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    glad to hear we've got the same answers,

    What did you put for q4 last part? I thought left tail didn't exist....well, i calculated that 0/12 you still accept H0. Therefore, i put that it's impossible to have left tail because you can't have negative no of packets or something like that.
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    need...s2....answers!!!

    i reckon i might have scraped an A, i bullsed up end of question 2 (such a retard not to set mean=variance even though i did initially then crossed it out). Everywhere else, although i was guessing at stuff, i reckon i was probably right. So A or B, but I really would like an A, don't want to blemish my record.

    I shouldnt have spent so much time on m2 - spent hardly any time revising for s2, BAH!
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    (Original post by Willa)
    need...s2....answers!!!

    i reckon i might have scraped an A, i bullsed up end of question 2 (such a retard not to set mean=variance even though i did initially then crossed it out). Everywhere else, although i was guessing at stuff, i reckon i was probably right. So A or B, but I really would like an A, don't want to blemish my record.

    I shouldnt have spent so much time on m2 - spent hardly any time revising for s2, BAH!
    AHHHHHHHHHHHHHHHHHHHH s2...Crappest exam in the world possibly?!
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    There was no left tail as the number was greater than 0.025. Did u reject H0 or accept H0?
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    (Original post by Mario_j)
    There was no left tail as the number was greater than 0.025. Did u reject H0 or accept H0?
    Oi Oi Mario! The lovely katie eh? (sorry, msn convo)! - Anyway, S2 is indeed the crappest exam of the year!
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    i accepted Ho. I'm pretty sure you have to accept Ho when the probability of getting the value is bigger than sign. level.

    I think it was 0.0687>0.025
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    good - if we've all done crap at s2, then hopefully you'll get marks added to normalise the results - yippy
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    (Original post by Willa)
    good - if we've all done crap at s2, then hopefully you'll get marks added to normalise the results - yippy
    I bloody well hope so! Still got P3 to do yet, and was hoping to do better in this so i wouldnt have to do so well in P3! Damn it!!!
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    I think i cocked up S1 . Oh well.
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    (Original post by maljosh)
    i accepted Ho. I'm pretty sure you have to accept Ho when the probability of getting the value is bigger than sign. level.

    I think it was 0.0687>0.025
    Didn't go too badly, much much much better than M1. In the last question I accepted Ho too. My answer to the last part was 28 packets, had to use logs. She needed to test 28 packets to get a critical region at the lower end. That is (I'm waffling i know) if she tested 28 packets and got no prizes you'd reject Ho and accept H1.
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    hi Jeremy, can you remember any of the answers from q2?
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    (Original post by JeremyG)
    Didn't go too badly, much much much better than M1. In the last question I accepted Ho too. My answer to the last part was 28 packets, had to use logs. She needed to test 28 packets to get a critical region at the lower end. That is (I'm waffling i know) if she tested 28 packets and got no prizes you'd reject Ho and accept H1.
    I got 17 :confused:
    The probability was 4/5 to get nothing, so
    (4/5)^x = 0.025
    x = ln(0.025)/ln(4/5) = 16.53

    Or did i do it wrong?
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    (Original post by maljosh)
    hi Jeremy, can you remember any of the answers from q2?
    I got
    i) 0.36
    ii) 0.504
    iii) 0.1152
    iv) 0.317
    v) 0.2
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    (Original post by maljosh)
    hi Jeremy, can you remember any of the answers from q2?
    Question 2.
    P for B =0.4
    P for A = 0.6

    0.064, same as you

    Find the probablilty of A winning one frame only and B winning
    then if A wins one frame
    ABBB
    BABB
    BBAB
    Probability for each combination = 0.0384
    multiplied by 3 gives : 0.1152

    Show the probability of B winning is 0.317

    AABBB
    ABABB
    ABBAB
    BAABB
    BABAB
    BBAAB
    Probability for each combination = 0.02304
    muliplied by 6 gives = 0.13804

    Add all three together, 0.064+0.1152+0.13804 = 0.317

    Just trying to remember the rest.............
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    Given that B wins, find the probability he wins in three frames:

    0.064/0.317 = 0.202
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    (Original post by JamesF)
    I got 17 :confused:
    The probability was 4/5 to get nothing, so
    (4/5)^x = 0.025
    x = ln(0.025)/ln(4/5) = 16.53

    Or did i do it wrong?
    Yep,

    0.8^x = 0.025
    xlog0.8 = log0.025
    x=log0.025/log0.8
    x= 16.53

    Not quite sure where I got 28 from......................
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    for part iv) i think you have to consider all possible winning situations like,

    BBB = 0.4^3=0.064

    for BBB only 1 possible way 3!/3!=1

    ABBB=(0.6*0.4^3)*3=0.1152

    for ABBB etc there are 4!/3! -1 = 3 ways


    AABBB etc=(0.6^2*0.4^3)*8=0.02304*6=0. 13824

    for AABBB= 5!/2!3! -4= 6 ways

    P(win)=0.064+0.1152+0.13824=0.31 744 or 0.317

    I think that's how i've done it....
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    for conditional probability i'm not quite sure what i've done...

    i'm sure i've used the correct formula... but can't remember getting 0.202,

    i think i got a slightly lower result... anyways, i'll get method marks,

    So for that paper i'll get at least a B.
 
 
 
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