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    (Original post by JasmineAladdin)
    P= 0.175 Q=0.15

    Or possibly the other way around
    I got q=0.15 and p=0.0175 or something like that
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    Actually quite enjoyed this morning ☺️ reckon the grade boundaries will be on the low side as well, which is a bonus.
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    Yeah I think that's right. I can't even remember exactly

    (Original post by Megan3245)
    I got q=0.15 and p=0.0175 or something like that
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    This is where everyone is hiding today
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    (Original post by Megan3245)
    I got q=0.15 and p=0.0175 or something like that
    The probabilities have to add up to 1.
    ~Therefore (0.5+0.15+(2x0.175)=1
    for yours (0.5+0.15+(2x0.0175)=0.685
    ( I Think this is right)
    Hope that helps you understand
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    (Original post by Megan3245)
    I got q=0.15 and p=0.0175 or something like that
    this is what I got too
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    (Original post by ODES_PDES)
    This is where everyone is hiding today
    lol yep
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    That's what I got

    (Original post by Hannaha124)
    For the Rebecca and Sarah q I got that Sarah wins when X= 3/2 and 2 so 2/5 and Rebecca's wins when X= -2 -1 and 1/2 so her probability is 3/5. Is this correct?


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    A lot of the questions (if not all!) were worth very low marks (mostly 2 - 3 mark questions). Stingy examiners. Oh well, that means if we get a few questions wrong, our overall marks wont be affected too much
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    HI does anyone know what they got for Sxy in question 3
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    (Original post by X_IDE_sidf)
    Thankyou, that mishap cost me around 2-4 marks
    That's a shame, but don't worry too much about it, the paper was very hard and the boundaries will be low.
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    (Original post by Megan3245)
    A lot of the questions (if not all!) were worth very low marks (mostly 2 - 3 mark questions). Stingy examiners. Oh well, that means if we get a few questions wrong, our overall marks wont be affected too much

    True i just hope i get a C or above. this was my retake got a u last year.
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    (Original post by CourtlyCanter)
    I said the Standard Deviation remained unchanged as the weight of the baby does not deviate from the mean weight of the babies.

    I also said the Mean remained unchanged as it is the same as the mean weight of the 50 babies before the addition of this new baby.

    Not sure if I got them right though. Someone please confirm or point out where I went wrong.
    I put that the mean remains the same, as you stated. However, I think that the standard deviation changes because if you look at the formula the Sum of x^2 values change and so does 'n' but the mean stays the same. If the calculation becomes different then the standard deviation would surely change?
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    (Original post by CourtlyCanter)
    I said the Standard Deviation remained unchanged as the weight of the baby does not deviate from the mean weight of the babies.

    I also said the Mean remained unchanged as it is the same as the mean weight of the 50 babies before the addition of this new baby.

    Not sure if I got them right though. Someone please confirm or point out where I went wrong.
    You are right mate!
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    (Original post by JasmineAladdin)
    That's what I got
    Rebecca doesn't win when x=-2

    ...

    OH **** **** **** **** ****
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    (Original post by JasmineAladdin)
    That's what I got
    Rebecca's does not win when the score is -1. It's a draw. 1/-1 = -1 hence they have the same score.
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    (Original post by C. Dixon)
    You are right in saying that the value doesn't deviate from the mean, however this means that the standard deviation will decrease as the data as a whole will deviate less. The mean will indeed stay the same.
    SD will not change! SD is root of sum of X - X bar divided by n. So as X- X bar is O is this case so sum of X-X bar do not change! so SD remains the same!
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    (Original post by uk_shahj)
    Wow I wrote the exact same points as you for both of them lol
    Yeah same here, i said the exact thing. Not sure if its write thou
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    Is there any unofficial mark schemes (with the questions) out there?
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    The mean would go up and the SD would change; the number was higher than de weight of most of the babies
 
 
 
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