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    Defo 3/38 for last question as teacher tells me. Unless my teacher is dumb, which is likely.
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    (Original post by Studious_Student)
    Did anyone else get for the outliers question that there will be some at the lower end and the upper end? Because 1.5 x 0.3 is 0.45 and - and + 17.9 is 17.45 and 18.35 respectively. So since the ranges are 17.4 for lower and 18.6 for upper there should be outliers at both ends right?


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    I'm afraid not, the lower quartile was 17.8 not 17.9, so 17.8 - (1.5 x 0.3) = 17.35. Lowest value was 17.4, so there are no outliers on the lower end. However, there are outliers on the higher end. But there is not enough data to be able to determine how many outliers there are.
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    (Original post by Duskstar)
    If 60-62 is an A, then a B is probably 53-55. However, I don't think boundaries are going to be that high this year - I'd say A is more likely to be around 58, and B respectively lower.
    hope so :s
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    On most previous mark schemes, if they don't to explain the H1, then they won't give marks for it even if you include it... Basically, if they don't ask for it, you don't need to put it.
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    (Original post by MathsMaestro)
    Do you have the mark scheme to prove me wrong? I've also said that my answer MAY be correct
    They've never had that in the mark schemes before, including 8 mark hypothesis testing questions from previous years.. Who are you?


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    (Original post by adrianblazesit)
    I'm afraid not, the lower quartile was 17.8 not 17.9, so 17.8 - (1.5 x 0.3) = 17.35. Lowest value was 17.4, so there are no outliers on the lower end. However, there are outliers on the higher end. But there is not enough data to be able to determine how many outliers there are.
    Damn, made a big mistake there ahaha, do you think I would at least get some method marks for the multiplication of 1.5 and 0.3?
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    So much beef in here lol sort it out in college tommorow
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    (Original post by Studious_Student)
    Damn, made a big mistake there ahaha, do you think I would at least get some method marks for the multiplication of 1.5 and 0.3?
    Maybe, it shows you knew what the formula was and you used it correctly (albeit with the wrong values), so I reckon 1 or 2 marks lost max.
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    (Original post by MathsMaestro)
    Do you have the mark scheme to prove me wrong? I've also said that my answer MAY be correct
    I'm backing jpg, you're wrong.
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    (Original post by adrianblazesit)
    I'm backing jpg, you're wrong.
    Adrianblazeit reply on Facebook


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    (Original post by adrianblazesit)
    Maybe, it shows you knew what the formula was and you used it correctly (albeit with the wrong values), so I reckon 1 or 2 marks lost max.
    Ahh ok, thank you very much, hopefully it's the former and not the latter, I want as many UMS as possible.
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    On the questions that said "find the expected value...", I multiplied the probability by the number but then rounded it because I didn't think you could expect to get a non-integer value - was I wrong to do this?


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    (Original post by henrygriff28)
    On the questions that said "find the expected value...", I multiplied the probability by the number but then rounded it because I didn't think you could expect to get a non-integer value - was I wrong to do this?


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    I think you only do that when youre calculating the most likely value
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    (Original post by Connorbwfc)
    It said 'He has to pick more than 2 before he gets one he likes' The only way this can happen is by getting a cherry first and second = 6/20 * 5/19 = 3/38.
    I think the last question said about him picking out more than 2 of the ones he doesn't like before he has a chocolate he does like.
    This means that he could have had 2 cherries, 3 cherries etc up to all 6 cherries before eating one that he likes.
    So you start with 6/20 x 5/19 x 14/18 (that's the probability of now picking one he likes) and then you do it again but with 3 cherries, then 4, 5, and 6
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    (Original post by jpetersgill)
    Swear it didn't ask why h1 had that form??


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    Same here I did not put the reason because it did not ask for it !
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    (Original post by DashDotDotDash)
    thank you ^^ (Sorry for the late reply,)
    Those videos will be useful for my other exams aas well, Thank you u w u
    you are very welcome ^^ ( don't worry it is okey )

    Good luck with your other exams
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    (Original post by MathsMaestro)
    This may be the correct answer. It is essentially P(X>2) = 1 - P(X<=2) = 1 - [P(X=0)+P(X=1)+P(X=2)]. PSC lol woohoo
    If that's the correct methodology, your method is wrong

    P(0) is just 14/20
    P(1) is just 6/20 * 14/19
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    (Original post by Duskstar)
    If that's the correct methodology, your method is wrong

    P(0) is just 14/20
    P(1) is just 6/20 * 14/19
    Even then, you can't do binomial distribution probablity unless you have a number of total trials, i.e (3C2) but there's no total, so it can't be done using P(X>....) etc
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    The only possible way, ignoring all other occurences aftwerwards, for him to have to take more than 2 before getting one he likes is simply not getting one he likes the first 2 attempts. End of, such a simple question. If he got Cherry Cherry other (p = 3/38) then you would say, "he had to take more than two chocolates before getting one he likes". Don't see what's so complicated about it tbh.
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    Hey

    I am just wondering how did u do the hypothesis test question !
    The thing that confused me, that there wasn't p=0.78 in the table so could not finish my answer

    Thank you
 
 
 
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