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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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    What do you think the grade boundaries will be like?
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    (Original post by Redblaziken8)
    What do you think the grade boundaries will be like?
    I don't think that they will be low.

    64 - A
    58 - B
    53 - C
    48 - D
    39 - E

    Just what I think anyway. Wasn't a hard exam, but I made a complete tit of it.
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    (Original post by Thequickspark)
    I don't think that they will be low.

    64 - A
    58 - B
    53 - C
    48 - D
    39 - E

    Just what I think anyway. Wasn't a hard exam, but I made a complete tit of it.
    that is so high
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    (Original post by Dapperblook22)
    I think I got 0.354, but there is usually a range of answers they accept.
    What was your method... I got an answer then divided by 6 😕
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    I hope you're wrong haha I messed up too on the 7 mark question I did sigma/sqrt(n) for standard deviation on the probability that each bottle and did standard deviation as as sigma for the mean question so I've probably lost 7 marks there
    Do they give ECF though?
    (Original post by Thequickspark)
    I don't think that they will be low.

    64 - A
    58 - B
    53 - C
    48 - D
    39 - E

    Just what I think anyway. Wasn't a hard exam, but I made a complete tit of it.
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    (Original post by priyanka pandya)
    that is so high
    As I said, don't worry as this is just based on my predictions and I am in no way competent enough to say that these are going to be accurate. If enough people found the exam hard, maybe they'll be significantly lower.
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    Here's a few of my answers:

    R=0.959
    Mode = 26
    IQR = 2

    The 2 things were mean and SD

    then Mean = 25.6
    SD= 5.26

    Equation of regression line was
    y=181+3.00x
    For the interpretation of gradient, i said that as the temperature increases by 1 degrees celcius, the mass of whatever dissolved etc increases by 3.00g
    the estimation was 385g
    i said large residuals so unreliable

    then the probability questions i got
    0.352
    0.136
    0.158
    0.792
    0.267
    and the 5 marker probability i got 0.0205

    The normal distrubuition ones i got
    0.941
    0.149
    0.792
    0.354
    0.993

    and the reduction in mean i got 2.3

    The 4 marker binomial one i got 0.891

    For the last question i disagree with first claim as after u make the new CI with the values in euros, 400 wasnt within the new CI, so claim was invalid.
    The second claim which working out the eurors to pounds conversion gave 166.67, and it wanted less than that, i got 14% and i said that therefore claim is not true.
    So first claim = invalid
    Second claim= Invalid

    and that's about it for all i remember/wrote down on extra paper i asked for.
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    (Original post by Thequickspark)
    I don't think that they will be low.

    64 - A
    58 - B
    53 - C
    48 - D
    39 - E

    Just what I think anyway. Wasn't a hard exam, but I made a complete tit of it.
    Those are a bit excessive. It wasn't easier than last year's.

    60 - A
    54- B
    48 - C
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    Highest I've seen for an A was 65
    (Original post by Thequickspark)
    As I said, don't worry as this is just based on my predictions and I am in no way competent enough to say that these are going to be accurate. If enough people found the exam hard, maybe they'll be significantly lower.
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    (Original post by OfficialChemist)
    Here's a few of my answers:

    R=0.959
    Mode = 26
    IQR = 2

    The 2 things were mean and SD

    then Mean = 25.6
    SD= 5.26

    Equation of regression line was
    y=181+3.00x
    the estimation was 385g
    i said large residuals so unreliable

    then the probability questions i got
    0.352
    0.136
    0.158
    0.792
    0.267
    and the 5 marker probability i got 0.0205

    The normal distrubuition ones i got
    0.941
    0.149
    0.792
    0.354
    0.993

    and the reduction in mean i got 2.3

    The 4 marker binomial one i got 0.891

    For the last question i agree with first claim as after u make the new CI with the values in euros, 400 was within the new CI, so claim was valid.
    However the second claim which working out the eurors to pounds conversion gave 166.67, and it wanted less than that, i got 14% and i said that therefore claim is not true.
    So first claim = valid
    Second claim= Invalid

    and that's about it for all i remember/wrote down on extra paper i asked for.
    What was your method for finding the reduction? That's my main issue so far. I got 1.7.

    The question was (just in case)

    what's the difference between the new and old mean to satisfy the requirement for only 10% of bottles to contain more than 1535ml. The original mean was 1525.
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    (Original post by jake4198)
    Those are a bit excessive. It wasn't easier than last year's.

    60 - A
    54- B
    48 - C
    I'm glad you disagree with me!
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    (Original post by Dynamic_Vicz)
    What was your method... I got an answer then divided by 6 😕
    I worked out the probabilty of 1 bottle containing the amount required, which is 0.841, so the probability of all 6 is 0.841^6 = 0.354
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    For the reduction of mean question I did it so z was 2.33 and got an answer of 12.4 how many marks would I lose cause wverything else was correct?
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    (Original post by CirclesAreRound)
    What was the last probability question even on about though?

    It said 4 people *not* from the sample of 500 so I was completely stumped as to where to get my probabilities from :-\

    I know that there were six combinations possible so the end answer had to be multiplied by 6/3! but apart from that colour me confused.

    I used the probabilities from the table regardless because I had no other option I could think of
    Yeah you'd use those probabilities and multiply by 3! to get 0.03 or something like that, I foolishly did multiplication by 2 and then 2! since I did them separately instead of doing the combinations for the total thing and got 0.0205 or something ;(
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    (Original post by Dapperblook22)
    I worked out the probabilty of 1 bottle containing the amount required, which is 0.841, so the probability of all 6 is 0.841^6 = 0.354
    I must've got my probability wrong. Hopefully, I'll still get marks for my method, which was the same as yours.

    I don't understand where I went wrong with finding the probability...
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    (Original post by Thequickspark)
    What was your method for finding the reduction? That's my main issue so far. I got 1.7.

    The question was (just in case)

    what's the difference between the new and old mean to satisfy the requirement for only 10% of bottles to contain more than 1535ml. The original mean was 1525.
    10% Contain more than 1535, so you're looking at the 90th percentile. Finding the Z value for this gives 1.28. You then sub this into the standardising formula to work out the new mean. The new mean ends up as being 1522.712. As it wants the difference, you do:

    1525-1522.712
    = 2.288
    =2.3
    i hope that's helpful g
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    (Original post by Parhomus)
    Yeah you'd use those probabilities and multiply by 3! to get 0.03 or something like that, I foolishly did multiplication by 2 and then 2! since I did them separately instead of doing the combinations for the total thing and got 0.0205 or something ;(
    Aww I did 4!.
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    (Original post by Thequickspark)
    I must've got my probability wrong. Hopefully, I'll still get marks for my method, which was the same as yours.

    I don't understand where I went wrong with finding the probability...
    Did you use the calculator or tables to find the probability?
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    (Original post by Dynamic_Vicz)
    What was your method... I got an answer then divided by 6 😕
    You get the probability it was greater than whatever value it was and do it to the power of 6
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    (Original post by OfficialChemist)
    10% Contain more than 1535, so you're looking at the 90th percentile. Finding the Z value for this gives 1.28. You then sub this into the standardising formula to work out the new mean. The new mean ends up as being 1522.712. As it wants the difference, you do:

    1525-1522.712
    = 2.288
    =2.3
    i hope that's helpful g
    Yeah, that's exactly what I would've done. I must've made a mistake on something alone the way. Thank you for explaining!
 
 
 
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