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OCR [MEI] Statistics 1 Exam Monday 25th Jan watch

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    Anyone just done this exam? I found it surprisingly easy. Section B was easier than the past papers I done as well, and the hypothesis testing was simple too! Which is good because I hate that.
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    Just me then? How did anyone else find it??

    By the way does anyone else find it annoying how small exam desks are. You have to fit a pencil case, exam question paper, exam answer booklet, graph paper, formula booklet and a calculator on a desk about the size of a food tray :confused:
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    It was ok, hypothesis testing is normally easy marks, so shame there wasnt more of it...
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    Was a pretty OK paper tbh, no weird stuff . "finished" like an hour early, then 20 minutes later realised I had: forgotten to actually draw the culmative frequency diagram, take into account there was different permutations of wet/dry/normal (whatever it was), and a couple of other things, oops. (luckily didn't ask if I could start my next exam until after I realised this haha)
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    I did two graphs, one on the booklet and a rushed one on graph paper. If one is wrong, which one would they mark?

    For the 'no student can win all three prizes', did anyone get 20*19*20 + 20*1*19 = 7980?

    For the 'comment on outliers' did anyone get 'about half of the lowest class and a small fraction of the highest one' which amounted to about a 7 outliers estimate in total?
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    for the no student can win all three prizes i did....20x19x19 + 20x20x19
    and for the outliers i said that you cant tell because we used a midpoint so we cant tell if there is any actual values.

    it was a gooood paper compared to the rest i thinkk 8-)
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    (Original post by sonal1066)
    for the no student can win all three prizes i did....20x19x19 + 20x20x19
    and for the outliers i said that you cant tell because we used a midpoint so we cant tell if there is any actual values.

    it was a gooood paper compared to the rest i thinkk 8-)
    First - We're agreed the first prize has 20 ways of doing it. For the second there are 19 ways of giving it to a student who didn't get the first prize and 1 way of giving it to the student who did. If the former then the last prize can go to anyone, if the latter then the last prize can go to anyone except the one who's received both. So 20x19x20 + 20x1x19. What reasoning led to yours?

    For outliers, maybe saying there could be some was enough, but I quantified how many there were likely to be and which classes they would be in.
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    i HATED the histogram question where you had to find the mean and standard deviation!!

    Also the prize one, i couldnt work out the answer to 'if no student can win all 3 prizes'...
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    @A100whoo

    I've just thought of a far simpler method of getting my answer to that. 20x20x20 minus 20 ways that a student can win 3 prizes. 8000 - 20 = 7980 as I got. So I'm now sure that's right.
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    (Original post by Game_boy)
    First - We're agreed the first prize has 20 ways of doing it. For the second there are 19 ways of giving it to a student who didn't get the first prize and 1 way of giving it to the student who did. If the former then the last prize can go to anyone, if the latter then the last prize can go to anyone except the one who's received both. So 20x19x20 + 20x1x19. What reasoning led to yours?

    For outliers, maybe saying there could be some was enough, but I quantified how many there were likely to be and which classes they would be in.

    well i havent got a clue.
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    (Original post by Game_boy)
    @A100whoo

    I've just thought of a far simpler method of getting my answer to that. 20x20x20 minus 20 ways that a student can win 3 prizes. 8000 - 20 = 7980 as I got. So I'm now sure that's right.
    ahh, thanks i get that now, shame its too late heh.
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    I messed up the all 3 prizes one, I can't remember but I knew it was wrong.
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    I did the PIN code question wrong- there were 10 digits, not 9! *facepalm
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    Yeah I reckon I got the prize one wrong (used combinations, as opposed to permutations) other than that pretty easy exam. Done trickier past papers anyway.
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    OH CRAP. How many marks was the pin code one in total? just realised I cocked that up majorly. Damnitdamnitdamnit.
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    (Original post by lowways)
    OH CRAP. How many marks was the pin code one in total? just realised I cocked that up majorly. Damnitdamnitdamnit.
    I heard that there was 2 for the first bit, which was the bit I screwed up. The next bit was 4!, methinks.
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    (Original post by morrison11)
    I heard that there was 2 for the first bit, which was the bit I screwed up. The next bit was 4!, methinks.
    I hope the first part was only 2 cuz that's what I got wrong too :] numbnuts here used 9999C4 (or 9999P4, can't remember which) instead of 10?4, if that makes sense. DOH!
    Oh well :]

    I also didn't know what a midrange was (1st question). My bad.

    Also, on the histogram question, what did everyone put as the skew?
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    (Original post by lowways)
    I hope the first part was only 2 cuz that's what I got wrong too :] numbnuts here used 9999C4 (or 9999P4, can't remember which) instead of 10?4, if that makes sense. DOH!
    Oh well :]

    I also didn't know what a midrange was (1st question). My bad.

    Also, on the histogram question, what did everyone put as the skew?
    For the pin one, i did 10^4 since you could have 10 digits at each of the 4 places. Then for part two it said he knew the 4 numbers but not their position, so i put must be 4!

    I think the histogram was a negative skew!
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    (Original post by LearningMath)
    For the pin one, i did 10^4 since you could have 10 digits at each of the 4 places. Then for part two it said he knew the 4 numbers but not their position, so i put must be 4!

    I think the histogram was a negative skew!

    I put negative first and then in a moment of idiocy which was masquerading as brilliant intelligence, I scribbled it out and put positive. Fml.


    But yeah, reckon I got the second part right cause it pretty much told me what to do ;D oh lord I have no hope :p:
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    Midrange is where you add together the lowest and highest value and then divide it by two.

    For the pin code on i did 1 over 10^4 and 1 over 4!, there were two marks for each.

    It was negatively skewed

    Also, the prize one, for the 'no student can win all three prizes' i did 20x20x19 + 20^3, because, for the first prize all 20 can get it and then the second any 20 can get it, now if the first two prices were won by the same person then the third can only be won by 19 of the people or it could happen that the first and second prizes were won by different people so the third could be won by any 20, hence 20^3.
 
 
 
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