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    Lost 16 marks C1
    Lost 14 marks C2
    How many marks can I lose in S1 to get an A?

    Roughly?
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    Can anyone post a link to last year's paper? It's not on the OCR website
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    (Original post by scrlk)
    IIRC Jan 2013 was by far the hardest paper they've done (49 for an A) so if you can do well on that I think you'll be prepared.
    You only needed 49 marks for an A? Surely that's way too low... Anyway I'm gonna try it now and see what I think
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    (Original post by Billy_Boi)
    Lost 16 marks C1
    Lost 14 marks C2
    How many marks can I lose in S1 to get an A?

    Roughly?
    It's normally similar grade boundaries to C2, so like 80% more or less for an A I guess, but it depends on how hard the paper is...
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    Does anyone know the next hardest paper after Jan 2013?
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    Go back to the one with the torn newspaper question, if i remember correct it was 2010.

    EDIT: It is Jan 2010
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    (Original post by LordHeskey)
    It's normally similar grade boundaries to C2, so like 80% more or less for an A I guess, but it depends on how hard the paper is...
    The S1 boundaries are usually a little bit lower than for C1 and C2 - usually about 75% for an A.
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    Past experience has shown that when seeds of a certain type are planted, on average 90% will germinate.A gardener plants 10 of these seeds in a tray and waits to see how many will germinate.
    (i) Name an appropriate distribution with which to model the number of seeds that germinate, givingthe value(s) of any parameters. State any assumption(s) needed for the model to be valid. [4]

    (ii) Use your model to find the probability that fewer than 8 seeds germinate. [2]

    Later the gardener plants 20 trays of seeds, with 10 seeds in each tray.
    (iii) Calculate the probability that there are at least 19 trays in each of which at least 8 seeds germinate.[4]

    Can anyone help with question (iii)????
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    (Original post by axle1234)
    Past experience has shown that when seeds of a certain type are planted, on average 90% will germinate.A gardener plants 10 of these seeds in a tray and waits to see how many will germinate.
    (i) Name an appropriate distribution with which to model the number of seeds that germinate, givingthe value(s) of any parameters. State any assumption(s) needed for the model to be valid. [4]

    (ii) Use your model to find the probability that fewer than 8 seeds germinate. [2]

    Later the gardener plants 20 trays of seeds, with 10 seeds in each tray.
    (iii) Calculate the probability that there are at least 19 trays in each of which at least 8 seeds germinate.[4]

    Can anyone help with question (iii)????
    part ii)

    X.B(10,0.9)
    P(X<8) = 0.072

    part iii)

    asking about probability of >8 seeds germinating, therefore 1-0.072=0.928

    So new distribution, Y.B(20,0.928)

    then asking P(Y>/=19) therefore P(X=19) + P(X=20)

    answer is 0.585
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    Can anyone help me with this question part (iv) ?

    On average, 25% of the packets of a certain kind of soup contain a voucher. Kim buys one packet ofsoup each week for 12 weeks. The number of vouchers she obtains is denoted by X.
    (i) State two conditions needed for X to be modelled by the distribution B(12, 0.25). [2]

    In the rest of this question you should assume that these conditions are satisfied.

    (ii) Find P(X ≤ 6). [2]

    In order to claim a free gift, 7 vouchers are needed.

    (iii) Find the probability that Kim will be able to claim a free gift at some time during the 12 weeks.[1]

    (iv) Find the probability that Kim will be able to claim a free gift in the 12th week but not before.[4]
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    (Original post by duncant)
    part ii)

    X.B(10,0.9)
    P(X<8) = 0.072

    part iii)

    asking about probability of >8 seeds germinating, therefore 1-0.072=0.928

    So new distribution, Y.B(20,0.928)

    then asking P(Y>/=19) therefore P(X=19) + P(X=20)

    answer is 0.585
    Thanks!! makes sense, i'm just over complicating stuff again ahaha. Do you by and chance know how to answer the last park of this question... (iii)

    Henry makes repeated attempts to light his gas fire. He makes the modelling assumption that theprobability that the fire will light on any attempt is 13.
    Let X be the number of attempts at lighting the fire, up to and including the successful attempt.

    (i) Name the distribution of X, stating a further modelling assumption needed. [2]

    In the rest of this question, you should use the distribution named in part (i).

    (ii) Calculate(a) P(X = 4), [3](b) P(X < 4). [3]

    (iii) State the value of E(X). [1](iv) Henry has to light the fire once a day, starting on March 1st. Calculate the probability that thefirst day on which fewer than 4 attempts are needed to light the fire is March 3rd. [3]

    Thanks again!
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    Did anyone find last year's paper? Please PM me if you'd rather not post it here.

    Thanks
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    (Original post by axle1234)
    Thanks!! makes sense, i'm just over complicating stuff again ahaha. Do you by and chance know how to answer the last park of this question... (iii)

    Henry makes repeated attempts to light his gas fire. He makes the modelling assumption that theprobability that the fire will light on any attempt is 13.
    Let X be the number of attempts at lighting the fire, up to and including the successful attempt.

    (i) Name the distribution of X, stating a further modelling assumption needed. [2]

    In the rest of this question, you should use the distribution named in part (i).

    (ii) Calculate(a) P(X = 4), [3](b) P(X < 4). [3]

    (iii) State the value of E(X). [1](iv) Henry has to light the fire once a day, starting on March 1st. Calculate the probability that thefirst day on which fewer than 4 attempts are needed to light the fire is March 3rd. [3]

    Thanks again!
    no problem, and I done this question about 20mins ago lol

    In b) we found P(X<4) which is what we need in iii). P(X<4)=19/27 (from b)).

    Now just think about it: he has to fail at lighting the fire less that 4 times the first day, the second day, but then succeed on the thrid day.

    therefore [1-(19/27)]^2 * 19/27 = 0.0618

    (failing twice then succeeding)
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    (Original post by Soraia)
    Did anyone find last year's paper? Please PM me if you'd rather not post it here.

    Thanks
    there you go good luck
    Attached Images
  1. File Type: pdf OCR+S1+June+2015.pdf (269.3 KB, 161 views)
  2. File Type: pdf OCR+S1+June+2015+mark+scheme.pdf (439.5 KB, 263 views)
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    (Original post by Dabo_26)
    there you go good luck
    That's great, thanks, that's my evening sorted!
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    (Original post by marioman)
    The S1 boundaries are usually a little bit lower than for C1 and C2 - usually about 75% for an A.
    Even better than I thought then To be honest that probably does reflect in the papers, although there are lots of very simple parts to stats they can ask some horrible questions
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    Hi,
    can anybody help me with question 5 ii and iii on the 2015 paper please.
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    does anyone know if we have to clear the memory from our calculators before every exam or if just once is enough for the whole exam period?
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    (Original post by dom117117)
    Hi,
    can anybody help me with question 5 ii and iii on the 2015 paper please.
    ii. Probability of a win = 0.27. There are 8 trials, so it's a binomial distribution.

    X ~ B(8, 0.27)

    P(X=2) = 8C2 * 0.27^2 * (1-0.27)^6
    = 0.309 (3 sf)

    iii. Using previous answer:

    [0.309 * 0.27] * [0.27 * (1-0.27)^2]

    Probability of success on 9th trial * probability of success on the 12th trial (i.e. 2 failures after the 9th trial and then a success).
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    (Original post by andy135)
    does anyone know if we have to clear the memory from our calculators before every exam or if just once is enough for the whole exam period?
    In theory you should clear it before every exam but no-one ever checks, unless you use a graphical calculator that can store text rather than just numbers.
 
 
 
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