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    Does anyone remember Q7 (ii) by any chance? ie the actual question. I don't really remember any of it :confused:
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    (Original post by Archer61)
    Substituting -a into x^4 gives you -a*-a*-a*-a giving you positive a^4.

    Also anyone remember how many marks question 3 was worth?
    Oh of course! Thank you i get it! How many marks do you reckon i lost for this? 2?
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    (Original post by AMDD)
    why would it not be (1/2a^3)*((1/4)a^4- - (1/4)a^4) inside the brackets which is where i seem to have gone wrong
    because (-a)^4 is just a^4 so it would be (1/2a^3)*((1/4)a^4 - +(1/4)a^4)
    In the variance calculations you have (-a)^5 which gives you -a^5 because of the odd power but in the E(X) you have an even.
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    (Original post by Archer61)
    Substituting -a into x^4 gives you -a*-a*-a*-a giving you positive a^4.

    Also anyone remember how many marks question 3 was worth?
    4+5+1 marks I think. Either 10 or 11
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    does anyone remember the order of accept/reject answers? wasn't able to store those numerical values onto my calculator, but I think I ended up getting something like accept, reject, accept?
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    (Original post by ResultInExcellence)
    does anyone remember the order of accept/reject answers? wasn't able to store those numerical values onto my calculator, but I think I ended up getting something like accept, reject, accept?
    the one involving the doctor was reject. the doctor was correct to say the mean has changed.

    the others were accepts i think, hope this helps
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    (Original post by ResultInExcellence)
    does anyone remember the order of accept/reject answers? wasn't able to store those numerical values onto my calculator, but I think I ended up getting something like accept, reject, accept?
    I assume you're talking about Q8 (ii)? If so I think all you need to do was model it using a binomial distribution

    EDIT: Nvm, misinterpreted your post
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    Does anyone know how to answer question 4???!!!! I thought it was impossible
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    (Original post by tokopoko)
    the one involving the doctor was reject. the doctor was correct to say the mean has changed.

    the others were accepts i think, hope this helps
    Yeah, I remember that one being reject as well. Well this seems to match so hopefully those are all correct!
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    (Original post by lllllllllll)
    I assume you're talking about Q8 (ii)? If so I think all you need to do was model it using a binomial distribution

    EDIT: Nvm, misinterpreted your post
    Aha sorry, to be fair, I did phrase it very badly, but couldn't think of a better way of doing so xD
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    pretty sure the probability is 0.6533 for 8ii), quite a few of my class got that, using different methods as well! Did anyone else get that?
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    What will be the grade boundary for an A this year do you think?
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    (Original post by lllllllllll)
    I think the continuity correction had to be -0.5 since we were finding P(X<x)
    I'm like 80% sure it was less than or equal to, not just less than.
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    Does anyone remember what they wrote for that 1 marker along the lines of whether it was necessary to have assumed that it was normally distributed?
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    (Original post by AMDD)
    What will be the grade boundary for an A this year do you think?
    i rekon itll be a=61 b=54 c=48 d=42 e=36 at a guess
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    * 8 ii) in full from memory* (checking my answer so might as well post it here)

    Find probability that out of 4 tests at least 1 results in a Type 2 error. Probability of p=0.3 or 0.5 or 0.7 is 1/3.

    P(type 2 when p=0.3) = 0
    P(type 2 when p=0.5) = 0.6047 (using tables) NB (from part i) we know critical region X >= 8
    P(type 2 when p=0.7) = 0.0933 (using tables)


    P(Type 2 error) = 1/3 x 0.6047 + 1/3 x 0.0933 = 0.23267

    B(4 , 0.23267)
    P(X at least 1) = 1 - P(X=0) = 1 - (1 - 0.23267)^4

    Ans = 0.6533
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    (Original post by irbif)
    i rekon itll be a=61 b=54 c=48 d=42 e=36 at a guess
    I think it'll be lower than that maybe 57 or 58 for an A - it was 59 last year and I'd say this paper was harder

    Posted from TSR Mobile
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    (Original post by NinjaPandaa)
    I think it'll be lower than that maybe 57 or 58 for an A - it was 59 last year and I'd say this paper was harder

    Posted from TSR Mobile
    ive got a feeling its gonna be high, thought with a lot of hypothesis testing will mean people will pick up a lot of marks
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    (Original post by ResultInExcellence)
    Does anyone remember what they wrote for that 1 marker along the lines of whether it was necessary to have assumed that it was normally distributed?
    n was large so central limit theorem applied
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    Hmmm, but then again a lot of people completely skipped out Q4 (10 marks) and Q8ii (5 marks)
    (Original post by irbif)
    ive got a feeling its gonna be high, thought with a lot of hypothesis testing will mean people will pick up a lot of marks
 
 
 
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