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    (Original post by gracebromby)
    I got 17 lol
    I got 17 too because-

    The given data tells us that there are 4 babies in the class ( 2w<3) with Width=1 Height=2.
    Therefore,
    2x1 = 2 cm^2
    So
    4/2= 2 (babies represented by each cm^2)

    We can also tell that Each cm (width) represents 1 kg.

    The class we are interested in finding the width and height of ( 3w<3.5)-
    Given 17= babies
    17/2= 8.5 cm^2

    Because Each cm (width) represents 1 kg, and our width is now 0.5 kg, the Width must be 0.5 cm.

    Therefore,
    8.5/0.5= 17 cm (Height)

    Anyone else agrees with this?
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    (Original post by anndz3007)
    For the diner question, do people add .405 and .450 and multiply by 77 or you find P( D intersect B interect R) add P( D intersect R intersect B' , i tried the second method and got 30 something but when i check up the diagram, the area of t is not counted, so i crossed it out and do the first method and got 68, im just so confused
    There were 77 people, but 40 were B and R, and 37 were B and not R. You do P(D|R intersect B) * 40 and P(D|R intersect not B) * 37. Then add those two values together. You end up with 33.
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    (Original post by Ppatel0)
    Omdz I mean the other way round but I can't remember the exact number of people they gave
    Yeah, I got 33 too.

    Although I believe it was (40 * 0.45) + (37 * 15/37) = 18 + 15 = 33
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    (Original post by tazza ma razza)
    well q4 was hard. Otherwise reckon I got 68/69+

    boundaries for 100, 90, 80 UMS?

    I think
    71+ for 100
    67+ for 90
    56+ for 80
    I'm thinking I got 64-68 ish, dunno. First solid A though, which is good news

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    (Original post by moshe015)
    Anyone get 0.65 for sarah and 0.2 for rebecca?

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    Yeah i did but i recon i was wrong😂😂😂
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    (Original post by CourtlyCanter)
    I got 17 too because-

    The given data tells us that there are 4 babies in the class ( 2w<3) with Width=1 Height=2.
    Therefore,
    2x1 = 2 cm^2
    So
    4/2= 2 (babies represented by each cm^2)

    We can also tell that the Each cm (width) represents 1 kg.

    The class we are interested in finding the width and height of ( 3w<3.5)-
    Given 17= babies
    17/2= 8.5 cm^2

    Because Each cm (width) represents 1 kg, and our width is now 0.5 kg, the Width must be 0.5 cm.

    Therefore,
    8.5/0.5= 17 cm (Height)

    Anyone else agrees with this?
    I agree, I did the exact same thing
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    I'm confused about how UMS works, if I get around 20 marks, I know that's a U but does that mean 0 UMS?
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    (Original post by X_IDE_sidf)
    Disagree with d i) \frac{0.42}{0.97} = 0.43299 (\frac{P(D\cap (B\cap R))}{P(B\cap R)} )
    My phone won't let me old half of your reply but here's why it was 0.45 and 15/37

    The probability of D given BnR is the proportion of BnRnD in BnR. Since from the diagram we can see BnRnD= 0.27 and BnR = 0.6, the answer is 0.27/0.6 which gives 0.45

    For probability of D given RnB' you do the same again. The total of RnB' is 0.15+0.22 = 0.37
    The portion of Rn'B which is also D is 0.15, hence the answer is 0.15/0.37 which is 15/37. This is definitely right because in the next question you use the fraction to work out that 15 out of 37 people have dinner at the hotel when they order a room but no breakfast.

    EDIT; I've spotted a couple of mistakes in your working. Firstly, P(Dn(RnB)) = 0.27, not 0.42, as it is the area in which all three circles intersect. Secondly, P(RnB) is 0.6, not 0.97. This is because RnB means the area inside B which is also inside R, which happens to be all of it (0.6). You've taken only the probability of R by the looks of it.
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    I got 0.36 for the last one, anyone else?

    Easy paper


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    Imo this paper was harder than last year's, I redid this module and I preferred last year's paper
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    (Original post by Bloom77)
    I got 0.36 for the last one, anyone else?

    Easy paper


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    Yeah I think that's what most people got.


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    When the extra value was added the pmcc would of moved closer to 0 right?
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    (Original post by hbaig27)
    If we got P and Q wrong, is the rest of the question wrong (Error carried forward) or will they give us method marks? This is for the Rebecca and Sara Q
    also wondering this.
    what did you get for p and q, out of interest?
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    Yeah
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    (Original post by Princepieman)
    I'm thinking I got 64-68 ish, dunno. First solid A though, which is good news

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    Well in Prince! Excellent stuff - no doubt if we both found it to be harder than expected, than boundaries will reflect and echo that Btw didn't realise you did edexcel - I'll PM you later this afternoon.
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    (Original post by adelemaexo)
    I'm confused about how UMS works, if I get around 20 marks, I know that's a U but does that mean 0 UMS?
    no, if you got a U your UMS will be somewhere between 0 and 40
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    (Original post by anna096)
    Yes thank you for explaining that! How many marks was this?
    only 2, dont stress
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    Did anyone get 0.680 for the standard deviation (where mean = 3.43) and 0.254 for the normal distribution question at the end of question 5? Some people seem to have similar answers but I might have got rounding errors.

    Also, what was the effect of the new data on the PMCC for question 3?
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    (Original post by fredhutchings)
    no, if you got a U your UMS will be somewhere between 0 and 40
    Ok brilliant thanks a lot
 
 
 
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