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    Yea I proved it. It was something like P(x>50) then continuity correction meant it was P(X>=50.5) i cant find greater than or equals to sign. stupid continuity corrections!!! i couldnt quite remember how to do it in exam tho so i did it lots of different ways until i got the right answer!
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    (Original post by starcrossed)
    yeh i proved that, i no all 3 types of approxs were iin the paper, wat other qs was hard, could everyone prve the long eqn, t4 + ............... etc
    Yeah you had to integrate between t0 and 0 to equal 0.95. then times by the fraction, then by 4 and you should get the same.
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    (Original post by Bhaal85)
    I actually worked out the X values for which the regions would be rejected. I know that it was definitely two tailed. Thereofre you should have 2.5% at either end.
    Thanks, that's what I did. I made a stupid mistake on question 2 and numbered the names 1-10, then got really confused. Maybe they'll got me a mark because they pity me.
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    i had the 0.918 for the X~p(7) thought it was to big so i crossed it out and did it using P(X<=3)...good bye 3 marks
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    (Original post by Bhaal85)
    Yeah you had to integrate between t0 and 0 to equal 0.95. then times by the fraction, then by 4 and you should get the same.

    i did all of that, except i didn't times by 4, lol i times by 250000 or something massive like that to get the same equation....
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    (Original post by Bhaal85)
    There was a hell of a lot of continutity(sp) corrections!!! anyone else prove that 0.231 thing?

    Yeah, I managed to prove it in the end. First time I forgot the cc, the next time I used the wrong one, then I finally got it right. Wasted a lot of time on that question.
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    All together it seemed a pretty good paper, not like all the higher pure modules, there didnt seem to be anywhere you could bin a load of marks, maybe just a few small things here and there. What do you reckon the grade boundaries will be like? Because i need a high UMS mark in this to get the grades for my oxford offer
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    (Original post by abc123)
    i did all of that, except i didn't times by 4, lol i times by 250000 or something massive like that to get the same equation....
    You should have taken out the multiplied the 0.95 by 62500, then the entire equation by 4.
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    (Original post by sock)
    Thanks, that's what I did. I made a stupid mistake on question 2 and numbered the names 1-10, then got really confused. Maybe they'll got me a mark because they pity me.
    I labelled them from 1-10 too, but then realised that I should have done 0-9, so I changed it, lol. It threw me quite a bit that question, how did people answer it? As this is something we don't normally do. I mean, what the hell, tables of random numbers? People pay for this crap? How does it work? lol. I then proceeded to select the '2nd' number from each set, until I had five.
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    (Original post by Bhaal85)
    You should have taken out the multiplied the 0.95 by 62500, then the entire equation by 4.
    yeah i see what your saying....although i'm quite sure the way i did would probably be valid since it all followed ok and i got to the final equation fine....i just multiplied the fraction (1/6750 or something) into the quartic, and then multiplied both sides by 250000 to get the right equation.
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    (Original post by Bhaal85)
    I labelled them from 1-10 too, but then realised that I should have done 0-9, so I changed it, lol. It threw me quite a bit that question, how did people answer it? As this is something we don't normally do. I mean, what the hell, tables of random numbers? People pay for this crap? How does it work? lol. I then proceeded to select the '2nd' number from each set, until I had five.
    I remember there being a question a bit like that in the mock exams in the textbook. I didn't understand it, so I asked a teacher about it, and he said chances are we wouldn't get a question like that, lol. Now I'm kicking myself for not asking him how to do it. Oh well, it's only 4 marks. Anyone else do M2 as well? What a fun afternoon that was, lol.
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    Oh well, THANK GOD for method marks!!!! lol. Anyway I really hope I get 90% plus.
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    I don't even know what a random no table is...I've never even heard of one before. I was so mighty confused when I got to that question. I just labeled each person with a number...in the order the numbers were given, then picked the first 5. tis about as wrong as you can get...but i figured doing something was better than nothign at all
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    Yeah, is that random number table stuff even on the syllabus, I've never seen it before, i just assigned each person with a range like 0 - 99 etc and then picked the first 5 numbers
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    yeah it is in the syllabus, its in the book somewhere in chapter 4 i think. its the bit with lots and lots of writing and u think, ill look at it later, and then dont. usually the question about number tables just come up in past papers about how u wud select a random sample, and u just vaguely mention it...not so this time
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    Well it was only 3 marks, and theres bound to be method marks in there somewhere, so whatever happened it cant be that bad
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    (Original post by starcrossed)
    how did everyone do q6b, did u get lamde as 1.05 and then use the poisson formula to find x=0, x=1 x=2 and then do 1 minus that answer

    I personally approximated them to a binomial distribution, with N 50 and P 0.021. Then did 1 - p(x = 1,2,3).
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    (Original post by pi is exactly 3)
    Well it was only 3 marks, and theres bound to be method marks in there somewhere, so whatever happened it cant be that bad
    to be honest i think the way you did it was correct anyway...its pretty simple when you think about it, its all about being confident witht the theory, even if its really simple...
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    (Original post by Bhaal85)
    I labelled them from 1-10 too, but then realised that I should have done 0-9, so I changed it, lol. It threw me quite a bit that question, how did people answer it? As this is something we don't normally do. I mean, what the hell, tables of random numbers? People pay for this crap? How does it work? lol. I then proceeded to select the '2nd' number from each set, until I had five.
    Label 1-10

    Divide the numbers you are given by 100. Round them UP to the nearest whole integer. Ignore repeats.

    Therefore 079 => 0.79 approx = 1 (Choose surname A)
    831 => 8.31 approx = 9 (Choose surname I)

    etc...
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    (Original post by Urdegish)
    I personally approximated them to a binomial distribution, with N 50 and P 0.021. Then did 1 - p(x = 1,2,3).
    I Used the binomial that way too, cant remember the answer though, was it something like 0.2 or 0.3?
 
 
 
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