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AQA M1SB 8th of june 2016 unofficial markscheme Watch

  • View Poll Results: [AQA MS1B 8th of June 2016] What raw mark do you think the A grade will be ?
    66
    20.00%
    65
    5.00%
    64
    10.00%
    63
    8.33%
    62
    13.33%
    61
    11.67%
    60
    15.00%
    59
    5.00%
    58
    11.67%

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    I used the Normal Stat Mode on the calculator and enter the following: Lower : 505 Upper: 99999999999999 s.d: 3.5 mean: 508.5 Which I get the probability of 0.841, then 0.841^6 which equals to 0.354 to 3sfg
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    I watched a video on using the calculator for this yesterday, but my calculator very kindly came up with math error (or some error) when it came to the exam, so I had to quickly do it manually which I probably messed up
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    (Original post by Cascadess)
    Don't worry Shakd.smxth I think that's the main thing people are arguing/confused about on this part. I multiplied by 4! because I thought that if you had A(1) and A(2) as the two different people of the first condition and B(1) and B(2) as the two different people of the second condition then there are four different events, which can happen in any order, and the way to find out the amount of ways things can be arranged is by factorial the quantity of items, which would be 4!

    But -jordan- Has told me that infact we should be multiplying by 4!/(2!*2!) which is 6.

    So i believe the full, correct, calculation is:

    (164/500)^2 * (121/500)^2 * 6

    We both got it wrong :/

    Yo i did exactly what u did but i multiplied it by 4 in the end. like (164/500)^2 * (121/500)^2 * 4

    How many marks will i get out of 5?!?!?!?!
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    (Original post by -jordan-)
    Oh well I think we'll all have scored some amount of marks, 2 or 3 maybe. Not sure how many I'll have lost but I guess it's an error they're anticipating people to make and will be reasonable with it.
    Sorry to bring this up again man, But i have now drawn out two trees, one dealing with each person as an individual condition, and one which has them as being the same. Like is my 24 tree just wrong? And if so could you explain again why?
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    (Original post by OfficialChemist)
    Yo i did exactly what u did but i multiplied it by 4 in the end. like (164/500)^2 * (121/500)^2 * 4

    How many marks will i get out of 5?!?!?!?!
    I recon you'd drop MAX two marks out of the 5, One for having a wrong answer at the end, and one for not properly working out how many different ways to arrange the selection of people. But if the mark scheme is nice, probably only drop 1!
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    (Original post by Cascadess)
    I recon you'd drop MAX two marks out of the 5, One for having a wrong answer at the end, and one for not properly working out how many different ways to arrange the selection of people. But if the mark scheme is nice, probably only drop 1!
    Alright thanks man respect
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    (Original post by Cascadess)
    Sorry to bring this up again man, But i have now drawn out two trees, one dealing with each person as an individual condition, and one which has them as being the same. Like is my 24 tree just wrong? And if so could you explain again why?
    I think you have to treat it as two probabilities and not four because four implies they are different events but they just have the same probability. You are looking for two events occurring twice.
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    (Original post by OfficialChemist)
    Alright thanks man respect
    Np Man
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    (Original post by -jordan-)
    I think you have to treat it as two probabilities and not four because four implies they are different events but they just have the same probability. You are looking for two events occurring twice.
    Damn, your case is very compelling, I think I may just be in denial about losing those marks Thanks a lot man
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    heres a poll to see how difficult other students found the exam:

    http://www.thestudentroom.co.uk/show....php?t=4150917
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    (Original post by Cascadess)
    Np Man
    I did that question like 4 times, examiner gona laugh at my paper lol
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    No sugarcoating or bias; was this paper harder than 2015 one?
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    (Original post by OfficialChemist)
    No sugarcoating or bias; was this paper harder than 2015 one?
    I don't think so
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    2 questions on how many marks I lose:
    For the 4 event probability I picked the wrong probability for one of the events but got the right one for the other and multiplied by 6?
    For the reduction of mean I used z as 2.33 to get 0.99 instead of 0.9 but I did everything else right?
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    (Original post by OfficialChemist)
    No sugarcoating or bias; was this paper harder than 2015 one?
    Nah
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    (Original post by Dapperblook22)
    I think I got 0.354, but there is usually a range of answers they accept.
    Does anyone remever part A of question 6.
    I remeber part B was the sample with 6 bottles but not sure what part A was.
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    (Original post by OfficialChemist)
    No sugarcoating or bias; was this paper harder than 2015 one?
    Yes I think 2015 was easier. However, I think I may have been a bit out of practise compared to the mock we did for the 2015 paper.
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    You wouldn't get a minus value as you would have to do the sd dividEd by root (n) which was 3.5/root (40) I think and you'd use that as your new sd, I ended up getting something like [257.9, 377.1]
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    (Original post by manutd1324)
    Does anyone remever part A of question 6.
    I remeber part B was the sample with 6 bottles but not sure what part A was.
    I can't really remember the context of the question, however the content of part A was just a standard normal distribution question.
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    (Original post by Dapperblook22)
    I can't really remember the context of the question, however the content of part A was just a standard normal distribution question.
    Ugh got them two bits completely wrong. I did the sample in part A and then just put that answer to the power of 6 for part B :no:. It was 7 marks too...
    So annnooyyeddd probably have no method marks for that either...
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    (Original post by SM-)
    2 questions on how many marks I lose:
    For the 4 event probability I picked the wrong probability for one of the events but got the right one for the other and multiplied by 6?
    For the reduction of mean I used z as 2.33 to get 0.99 instead of 0.9 but I did everything else right?
    for the 4 probability one probably you probably got 3 outta 5


    and you'll probably lose two marks for the second one as well

    For both of those, one for the mistake lost and one mark for the bad answer at the end you would have got.
 
 
 
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