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    (Original post by kprime2)
    Yes



    I don't have the paper yet
    Alright Thanks anyway
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    (Original post by Rafeerahman)
    in the skewness question, did anyone write that the mean (3.43) was close to the median ( 3.47) so normal distribution can be used. i wanted to do the 3 (mean-median)/sd calculation but later i didnt do that uggghhhhh. i also wrote that the standard deviation he used (0.65) was a bit deviated from the standard deviation we had to find (0.68). i said despite these small inaccuracies, a normal distribution was fairly appropriate. am i getting 0 out of 3 for this?
    I reckon you'll be able to say that they are close, so a normal distribution is appropriate. That's usually only 2 marks though. What worries me is that I didn't mention anything about the probability, P(W<3), which I think we were meant to include. I don't know what to say about that, though.

    The actual data gives P(W<3)=0.18, which isn't very well supported by the model. (0.2546).
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    (Original post by fowlerj)
    I reckon you'll be able to say that they are close, so a normal distribution is appropriate. That's usually only 2 marks though. What worries me is that I didn't mention anything about the probability, P(W<3), which I think we were meant to include. I don't know what to say about that, though.

    The actual data gives P(W<3)=0.18, which isn't very well supported by the model. (0.2546).
    They did they said use your answer to (e) and (f) or whatever it was , one of the parts it asked you to include was the question where you were asked find P(W<3)
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    (Original post by HaveFa1th)
    i don't suppose you would get any credit for saying "little to no skew" so normal distribution is a good decision and data measured is continous?
    (Original post by Rafeerahman)
    in the skewness question, did anyone write that the mean (3.43) was close to the median ( 3.47) so normal distribution can be used. i wanted to do the 3 (mean-median)/sd calculation but later i didnt do that uggghhhhh. i also wrote that the standard deviation he used (0.65) was a bit deviated from the standard deviation we had to find (0.68). i said despite these small inaccuracies, a normal distribution was fairly appropriate. am i getting 0 out of 3 for this?

    (Original post by fowlerj)
    I reckon you'll be able to say that they are close, so a normal distribution is appropriate. That's usually only 2 marks though. What worries me is that I didn't mention anything about the probability, P(W<3), which I think we were meant to include. I don't know what to say about that, though.

    The actual data gives P(W<3)=0.18, which isn't very well supported by the model. (0.2546).
    5(e) I didn't realise it said use answer to part d too. In part d P(W<3) = 0.2546 whereas the actual P(W<3)=0.18. My comment would be that this just further consolidates that the normal model is not really accurate. Also, I don't think saying the decision to use the normal model on the basis that the median and median are close will be accepted. While the median and mean are close, the data is not nearly symmetrical, if it were part d P(W<3) would have yielded a more accurate approximation.
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    Guys, for the first question, I got the right answer: 5.69 but then I rounded saying it was 5 (Whole wiggles). Reckon I'd lose a mark or something?

    Also, what do you reckon the grade boundaries will be?
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    (Original post by kprime2)
    Wouldnt be surprised
    Really, can it fall as low as 55?
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    (Original post by SANTR)
    Really, can it fall as low as 55?
    Well I thought this was quite a tricky paper. But it seems that the students found it easier than I think, so I think the boundary would be 56-57 for A.
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    (Original post by kprime2)
    Well I thought this was quite a tricky paper. But it seems that the students found it easier than I think, so I think the boundary would be 56-57 for A.
    What's your predictions for C1?
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    (Original post by SANTR)
    What's your predictions for C1?
    I would say about 58-59
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    (Original post by kprime2)
    Definitely
    Hi, I got the following:
    C1: 67 marks
    C2: 68 marks
    S1: 56 marks

    Would that be an A overall?
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    (Original post by PMONY)
    Hi, I got the following:
    C1: 67 marks
    C2: 68 marks
    S1: 56 marks

    Would that be an A overall?
    Yes it is
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    (Original post by kprime2)
    Well I thought this was quite a tricky paper. But it seems that the students found it easier than I think, so I think the boundary would be 56-57 for A.
    Confirmed - 56 for an A.

    http://qualifications.pearson.com/co...ries-Final.pdf
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    so what would 90 UMS be bro and what would 85 UMS be too
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    so what would 90 UMS be bro and what would 85 UMS be too
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    (Original post by ALPACINO357)
    so what would 90 UMS be bro and what would 85 UMS be too
    85 UMS is around 60/75 I think. 90 is about 63/75.
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    (Original post by kprime2)
    Soz for the delay. Got caught up with the Euro games

    I thought this was quite tricky
    https://drive.google.com/open?id=0B3...GhaRHh2MHlnaDQ

    EDIT: 5(e) I didn't realise it said use answer to part d too. In part d P(W<3) = 0.2546 whereas the actual P(W<3)=0.18. My comment would be that this just further consolidates that the normal model is not really accurate. Also, I don't think saying the decision to use the normal model on the basis that the median and median are close will be accepted. While the median and mean are close, the data is not nearly symmetrical, if it were part d P(W<3) would have yielded a more accurate approximation.

    Hello,
    Thank you for this and would you be able to possibly explain the working out for question 2e.

    Thank you!
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    (Original post by Rubydiamond3131)
    Hello,
    Thank you for this and would you be able to possibly explain the working out for question 2e.

    Thank you!
    R = 1/X

    the probabilities remain the same so multiply each new R by each probability
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    (Original post by kprime2)
    R = 1/X

    the probabilities remain the same so multiply each new R by each probability
    Sorry - I don't understand where the -1/2 is from..
    I am very sorry to bother you, I just needed some help!

    Did you multiply the x value with the X value and then do 1 divided by this...
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    (Original post by Rubydiamond3131)
    Sorry - I don't understand where the -1/2 is from..
    I am very sorry to bother you, I just needed some help!

    Did you multiply the x value with the X value and then do 1 divided by this...
    1/(-2) = -1/2

    R values multiplied by probabilities
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    (Original post by kprime2)
    1/(-2) = -1/2

    R values multiplied by probabilities
    Will you be doing model answers for C3 C4 and M1 this year
 
 
 
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