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    (Original post by Tick7)
    I hate to break it to you but s/he is right as they've multiplied (5/12 x 5/12 x 5/12 x 7/12) by 4C3 because there a 4C3 ways in which you can order 4 items in which 3 are the same
    yahh, anoo i found out like 3 hours ago but i will only loose 2 out of 4 so its fine
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    (Original post by zangorou)
    I got Q2 right?!?! I think I'm going to pee.



    I don't think an A will be any higher than that. January 2013 was 57 and that was the easiest S1 paper I completed.

    January 2013 was soooooo easy. I can only see question 2 where I've messed up, but should get some method marks. Maybe now I can stop worrying :O
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    (Original post by jelmes96)
    January 2013 was soooooo easy. I can only see question 2 where I've messed up, but should get some method marks. Maybe now I can stop worrying :O

    Nice to know someone enjoyed that exam...
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    i'm so gutted. i thought i had done so well in this exam but looking at the mark scheme, i have made one HUGE error.


    I think i've got 100% up until Q7. But in Q7, starting with the tree diagram, i did every branch where it was meant to be 0.05 as 0.5. I forgot the extra 0 !!!! This meant that i got the remaining final answer parts of Q7)ii)A)B)C)iii) wrong!!! HOWEVER, i am pretty confident that i did all the correct workings out and all the correct methods, just with using 0.5 instead of 0.05.
    I got the conditional probability and the last question right in my workings, and i definitely did with Q7)ii)A)B), i just simply used 0.5 where it should have been 0.05

    I suppose my workings and answers clearly show i know how to do the questions, i just used the wrong value

    How many marks am i looking at dropping with this mistake?
    The last question was out of 18 and i think i'll probably get -2 for the tree diagram, then -2 for ii)A) and then -3 for ii)B) and then -2 for ii)C) and then -2 for iv)

    owh i was hoping for 100UMS

    will i even get an A do you think?
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    (Original post by RomanMarkOliver)
    i'm so gutted. i thought i had done so well in this exam but looking at the mark scheme, i have made one HUGE error.


    I think i've got 100% up until Q7. But in Q7, starting with the tree diagram, i did every branch where it was meant to be 0.05 as 0.5. I forgot the extra 0 !!!! This meant that i got the remaining final answer parts of Q7)ii)A)B)C)iii) wrong!!! HOWEVER, i am pretty confident that i did all the correct workings out and all the correct methods, just with using 0.5 instead of 0.05.
    I got the conditional probability and the last question right in my workings, and i definitely did with Q7)ii)A)B), i just simply used 0.5 where it should have been 0.05

    I suppose my workings and answers clearly show i know how to do the questions, i just used the wrong value

    How many marks am i looking at dropping with this mistake?
    The last question was out of 18 and i think i'll probably get -2 for the tree diagram, then -2 for ii)A) and then -3 for ii)B) and then -2 for ii)C) and then -2 for iv)

    owh i was hoping for 100UMS

    will i even get an A do you think?
    There was a section b probability question in the June 2010 paper and the examiner briefly commented on cases where candidates gave the wrong probability in the diagram.
    Basically , although this lost most of the marks for the diagram , the report states a generous follow through was in place for the later questions and even went up to 3/4 marks (losing a1 only). While its impossible to predict exactly how they will mark it this time you should definitely pick up some method marks if you did the right working with the wrong numbers, as it were.


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    I think I got 56/57 out of 72.... possibly an A or is it more like to be a B :/
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    (Original post by profit_master3)
    Most "UPTO DATE" UNOFFICIAL MARKSCHEME:

    (post by profitmaster_3)

    UNOFFICIAL MARK SCHEME


    Please FEEL free to Update and Approve and most of all discuss. This is just the most s1 Answer scheme agreed upon the the student room community.

    1i) mean = 249.4, standard deviation = 14.5 [3]
    ii) new mean = 209.46, new standard deviation = 13.0 [3]

    2i) P(2 women, 1 man) = (5C2x5C1)/10C3 = 5/12 [4]
    ii) P(four evenings with 2 women, 1 man) = (5/12)^4 + 4C3(5/12)^3(7/12) = 0.1989 [4]

    3i) P(5underweight bags) = 50C5(0.1)^5(0.9)^45 = 0.1849 [3]
    ii) P(at least one underweight bag in 20) = 1-(0.9)^20 = 0.8784 [2]
    iii) E(X) = np = 48x0.8784=42.2 [2]

    4i) 3! x 1/6 x 1/5 x 1/4 = 1/20 [2]
    ii) E(X)= sum xP(X=x) = £1507.50, Var(X)=E(X^2)-E(X)^2 = £445511.25 [5]

    5i) Because the researcher has assumed that the probability that a randomly selected person makes a correct identification is 0.5, as she assumes that there is a 50:50 chance of guessing the difference between tap and bottled water [2]
    ii) Because the point of the test is to determine if the probability that a randomly selected person makes a correct identification is in fact greater than 0.5 [1]
    iii) P(X≥13)=1-P(X≤12)=1-0.8684=0.1316>0.05 Therefore, 13 is not a member of the critical region and it is not significant. We accept H0. There is insufficient evidence from the data to suggest the probability that a randomly selected person makes a correct identification is greater than 0.5 [5]

    6i) Median=3.32, LQ=(2.82+2.84)/2=2.83, UQ=(3.70+3.72)/2=3.71. IQR=0.88 [3]
    ii) Draw box and whisker plot [3]
    iii) 1step = 1.5xIQR=1.32. UQ+1.32=5.03, therefore no outliers above. LQ-1.32=1.51. Therefore 1.39 is an outlier below. It shouldn't be removed as there is no evidence to suggest that it forms a separate pool of data or is not a genuine data item. [4]
    iv) Median=3.5. UQ=3.84, LQ=3.13. IQR=0.71 [3]
    v) Males have a higher average weight (higher median). Males have less variation in birth weight (lower interquartile range). [2]
    vi) (8/200)x(7/199). You can probably use any values from 7-10/200 (it depends how you read the graph). [3]

    7i) Draw probability tree diagram [4]
    iiA) 1-0.9x0.95x0.95=0.1878 (4 d.p.) [3]
    iiB) 0.1x0.8x0.95+0.9x0.05x0.8+0.9x0. 95x0.05=0.1548 (4 d.p.) [4]
    iii) 0.1548/0.1878=0.8242 [3]
    iv) P(Hit,Hit,Hit) + P(Miss,Miss,Miss,Hit,Hit,Hit) = 0.0056 (4 d.p.) [4]

    Paper to be uploaded SOON.
    is this not my post from earlier today?
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    Hey for example for question 2, If you got the wrong value for part 1 would you get error carried forward marks for part 2???? because otherwise for 1 silly mistake you lose 8 marks!!!! could someone please reply as im very worried about this
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    Well I just realised I accidentally typed in a 3 instead of a 4 to work out something for Q2 part i) which would mean I also got part ii) wrong.. And to make matters worse I just found out I misread the last question and messed that up. Ugh only have myself to blame
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    Does anyone know if we get error carried forward marks!!!
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    (Original post by Law-Hopeful)
    Have to multiply by 4! because there are 4! ways that (5/12)^3 x (7/12) could occur (order wise). :/
    Ah I got the answer, but did it a different way 0.O
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    (Original post by Law-Hopeful)
    You actually had to multiply it by 4C3 not 4!... I completely omitted that Hopefully only lose 1 mark but probably 2 :/
    How did you find the exam?
    Hmm thats what I thought.

    Oh no :/

    I thought it was a pretty nice exam actually, bar the last questions How about you?
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    (Original post by olakase)
    Well I just realised I accidentally typed in a 3 instead of a 4 to work out something for Q2 part i) which would mean I also got part ii) wrong.. And to make matters worse I just found out I misread the last question and messed that up. Ugh only have myself to blame
    I did the last question wrong three times because I misread it each time...eventually got what I think is the correct answer but who knows.
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    everyone on this **** website only cares about themselves, DO WE GET ERROR CARRIES ****ING FORWARDS!!!!!!!!!!!
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    (Original post by Felttip)
    everyone on this **** website only cares about themselves, DO WE GET ERROR CARRIES ****ING FORWARDS!!!!!!!!!!!
    You would probably get method marks.
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    (Original post by Felttip)
    everyone on this **** website only cares about themselves, DO WE GET ERROR CARRIES ****ING FORWARDS!!!!!!!!!!!
    You wouldn't get FT marks as a matter of principle, carried forward marks are decided once the papers are in and the examiners can see how many people have made the error. If it is a common mistake then it is likely that you will get some method marks and in special cases even all method marks or full marks , this is shown on the mark scheme under the code SC and is usually given when an abnormally large number of candidates made the same mistake.


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    Just out of interest, how many on here did this during their AS year, and how many during their A2 year?
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    (Original post by pak1994)
    I did the last question wrong three times because I misread it each time...eventually got what I think is the correct answer but who knows.
    Haha well done to you! I know I've definitely got it wrong


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    What did people put for the reasons why the alternative and null hypothesis were used?
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    (Original post by coeeey9)
    What did people put for the reasons why the alternative and null hypothesis were used?
    A whole load of mumbling rubbish...
 
 
 
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