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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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    For the 4 mark binomial question where it was X ~ B(50,0.8) for p(35<X<45) the standard method is to change it to X' ~ B(50,0.2) and change the ranges. However, I used the graphical calculator to just work out X ~ B(50,0.8) for p(35<X<45) and got the same answer as OP says. Anyone think I will lose marks here?
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    (Original post by 1z3)
    For the 4 mark binomial question where it was X ~ B(50,0.8) for p(35<X<45) the standard method is to change it to X' ~ B(50,0.2) and change the ranges. However, I used the graphical calculator to just work out X ~ B(50,0.8) for p(35<X<45) and got the same answer as OP says. Anyone think I will lose marks here?
    I doubt it - I think there has been a case before where they accepted an answer for binomial directly from the calculator where p > 0.5.
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    (Original post by Dapperblook22)
    I can't remember the exact question, so I will try to explain as best as I can.

    There were 4 outcomes, two pairs of which were the same. As you were estimating from the sample, you divide each outcome from the total in the sample (500). Multiply them together, then multiply by 6 as this was the number of ways of arranging them.

    This came down to the following calculation:

    {[(176-28)/500]^2} x {[(140-6-3)/500]^2} x 6 = 0.030786 = 0.0308

    Sorry this is a general answer and not specific, however I have forgotton most of the wording of the question
    Great cheers, I multiplied by 16 instead of 6, should still get some method marks
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    What about question 3b? I got 0.007.... or something less than 0.01.

    And the last question about the claims? The first claim is valid or not?? I just directly multiplied the c.i by 1.20 and said it is valid because 400 is within the new c.i. But some students calculate the new c.i using the mean+-standard error.
    The second claim said at most 25% and i got 17.5%, so I said it is invalid because it can not get to 25% as the highest is 17.5%.
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    (Original post by JPencil)
    Anyone else leave all of the probabilities in fractional form...?

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    don't worry I did this in my mock and also got a bunch of other questions wrong and still got an A


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    In the question where it gave 3 new values to calculate a mean etc, the question only had 2 unknown vales, can anyone tell me what they did there?


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    My answers (no idea if they are right ir not just contributing):

    1.)a.) r=0.9591...
    b.) Strong correlation blah blah

    2.)a.)i.) 26
    ii.) Median: 25, IQR: 2
    b.) Mean and range
    c.) Mean: 25.6, Range: 45

    3.)a.)i.) 0.352
    ii.) 0.136
    iii.) 0.158
    iv.) 0.792
    v.) 0.267
    b.) 0.00513

    4.)a.) y= 181.3 + 3.0042x
    b.) (context of gradient)
    c.) 385
    d.) I put the estimate was accurate as even though the residuals are large so are the values of Y but this could well be wrong.

    5.)a.)i.) 0.941
    ii.) 0.149
    iii.) 0.792
    iv.) Unity
    b.) 1522.7
    c.)i.) 0.355
    ii.) 0.993

    6.)a.)i.) 0.0262
    ii.) 0.880
    iii.) 0.101
    iv.) 0.891
    b.) Mean: 54, Var: 44.28

    7.)a.) (257.92,377.08)
    b.)i.) 381 not equal to 400 so not valid
    ii.) 7/40 not equal to 0.25 so not valid (which is wrong, it IS valid but i didnt read the question properly)

    Hope this helps, please point out if I went wrong somewhere which I'm sure i did in multiple places.
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    Full ums prediction?
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    (Original post by asdfaeth)
    In the question where it gave 3 new values to calculate a mean etc, the question only had 2 unknown vales, can anyone tell me what they did there?


    Posted from TSR Mobile
    The groups where <20 or >30 which had frequencies of 1 and 2 respectively meaning 3 in total.
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    (Original post by 1z3)
    For the 4 mark binomial question where it was X ~ B(50,0.8) for p(35<X<45) the standard method is to change it to X' ~ B(50,0.2) and change the ranges. However, I used the graphical calculator to just work out X ~ B(50,0.8) for p(35<X<45) and got the same answer as OP says. Anyone think I will lose marks here?
    I kinda did the same - was running out of time and couldn't work out how to do it, so I just used the summation function on my calculator xD
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    (Original post by asdfaeth)
    In the question where it gave 3 new values to calculate a mean etc, the question only had 2 unknown vales, can anyone tell me what they did there?


    Posted from TSR Mobile
    Each of the 3 x values had a y value of 1 because you had to spit the last Unknown into two values as there was only 2
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    (Original post by xs4)
    I apologise I do not remember many questions by this is what I have thus far. Contributions towards this and corrections would be welcomed - Thank you to everyone that contributed

    AQA MS1B Unofficial Mark Scheme

    Q1a) PMCC (r) calculation [3 marks]
    r = 0.95(9..)
    b) Interpret your value for r [2 marks]

    Q2a i) Mode = 26 [1 mark]
    ii) mean and range or SD [2 marks]
    b) find the mean and range [3 marks]
    mean = 25.6
    range = 45

    Q3) probabilities - different ages and bands 1-3
    ai) 0.136
    ii) 0.158
    iv) 0.729
    v) I can't remember [9 marks for part a]
    b) 0.0308 [5 marks]

    Q4) Normal distribution
    a) P(X<1540)
    ii) P(X<1535?)
    b) P(1515?<X<1540)

    Q5)Binomial distribution - loom bands
    ai) P(Red loombands = 2)
    (50C4) * (0.18)^4 * (0.82)^46 = 0.0262


    ii) P( Y≤10) Calculate the probability at most 10 were yellow

    Therefore X ~ B(50, 0.15)

    P(Y≤10) = 0.8801


    iii) P(green and purple< bands which gave you

    X ~ B(50, 0.5)
    Binomial

    aiii) Something to do with green and purple bands which gave you

    X ~ B(50, 0.5)
    iv) P(35<W'&Y'<45)
    =>X ~ B(50, 0.8)

    For P(35 < X < 45)

    (Therefore using the complementary event denoted by Y )

    Y ~ B(50, 0.2)

    P(6 ≤ Y ≤ 14) = P(Y≤14) – P(Y≤5) = 0.8913

    b) Calculate the mean and variance for 300 red loom bands

    μ = np → 300 * 0.18 = 54

    σ² = np(1–p) → 54(1–0.18) = 44.3

    Q6) Normal distribution
    volume of the bottles in a pack of 6
    mean 508.5 , SD= 9.6?
    ai) P(X>505) [NOT SURE IF THIS IS ACTUALLY ai)]


    Q7) confidence intervals people buying euros
    a) calculate the limits to 2 dp
    b) comment on the claim(s) that [3 marks]
    i) ?
    ii) 25% of the customers bought fewer than 200 euros
    first claim invalid
    2nd claim customers bought less than 200 euros
    after the conversions of £ to euros and the calculation (no. of customers/n) this claim came to be valid

    - I'm unsure of the exact percentage
    I got 0.0409 for part 3(b)
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    (Original post by JamieOH)
    My answers (no idea if they are right ir not just contributing):

    1.)a.) r=0.9591...
    b.) Strong correlation blah blah

    2.)a.)i.) 26
    ii.) Median: 25, IQR: 2
    b.) Mean and range
    c.) Mean: 25.6, Range: 45

    3.)a.)i.) 0.352
    ii.) 0.136
    iii.) 0.158
    iv.) 0.792
    v.) 0.267
    b.) 0.00513

    4.)a.) y= 181.3 + 3.0042x
    b.) (context of gradient)
    c.) 385
    d.) I put the estimate was accurate as even though the residuals are large so are the values of Y but this could well be wrong.

    5.)a.)i.) 0.941
    ii.) 0.149
    iii.) 0.792
    iv.) Unity
    b.) 1522.7
    c.)i.) 0.355
    ii.) 0.993

    6.)a.)i.) 0.0262
    ii.) 0.880
    iii.) 0.101
    iv.) 0.891
    b.) Mean: 54, Var: 44.28

    7.)a.) (257.92,377.08)
    b.)i.) 381 not equal to 400 so not valid
    ii.) 7/40 not equal to 0.25 so not valid (which is wrong, it IS valid but i didnt read the question properly)

    Hope this helps, please point out if I went wrong somewhere which I'm sure i did in multiple places.
    I think you either missed out the question that you had to find the reduction in the value of mean or you got that wrong?
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    I'm pretty sure both of the comparisons you had to make at the end were reasonable or seemed likely. The first you had to convert 400 euros into pounds. This was inside the confidence interval so it seems likely that the mean is that.

    The second there was 8/40 that met the criteria giving a percentage of 20% which meant that the claim it was at most 25% also seemed likely.
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    (Original post by Citric_Xenon)
    I think you either missed out the question that you had to find the reduction in the value of mean or you got that wrong?
    Must have missed it. Oops.
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    (Original post by JamieOH)
    My answers (no idea if they are right ir not just contributing):

    1.)a.) r=0.9591...
    b.) Strong correlation blah blah

    2.)a.)i.) 26
    ii.) Median: 25, IQR: 2
    b.) Mean and range
    c.) Mean: 25.6, Range: 45

    3.)a.)i.) 0.352
    ii.) 0.136
    iii.) 0.158
    iv.) 0.792
    v.) 0.267
    b.) 0.00513

    4.)a.) y= 181.3 + 3.0042x
    b.) (context of gradient)
    c.) 385
    d.) I put the estimate was accurate as even though the residuals are large so are the values of Y but this could well be wrong.

    5.)a.)i.) 0.941
    ii.) 0.149
    iii.) 0.792
    iv.) Unity
    b.) 1522.7
    c.)i.) 0.355
    ii.) 0.993

    6.)a.)i.) 0.0262
    ii.) 0.880
    iii.) 0.101
    iv.) 0.891
    b.) Mean: 54, Var: 44.28

    7.)a.) (257.92,377.08)
    b.)i.) 381 not equal to 400 so not valid
    ii.) 7/40 not equal to 0.25 so not valid (which is wrong, it IS valid but i didnt read the question properly)

    Hope this helps, please point out if I went wrong somewhere which I'm sure i did in multiple places.
    I don't remember getting unity as an answer anywhere, was this in the normal distribution question?
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    What was the necessary assumption for q6 b
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    (Original post by Dapperblook22)
    I don't remember getting unity as an answer anywhere, was this in the normal distribution question?
    Yes, it was the last one before you had to find the new value of mu.
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    (Original post by -jordan-)
    Yes, it was the last one before you had to find the new value of mu.
    ah, I must have missed it . At least it is only 1 mark.
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    (Original post by mcride98)
    What was the necessary assumption for q6 b
    I said that the probability of the volume of water in each bottle was independent... but now I'm thinking it was simply that each bottle in the pack was the same size.
 
 
 
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