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Official Edexcel S3 thread Wednesday 20 May 2015 Morning [6691] Watch

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    A bit tricky, but generally I was happy when I walked out of the room.

    Was it necessary for the weight loss distribution to be normal? I answered no.
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    (Original post by simonli2575)
    A bit tricky, but generally I was happy when I walked out of the room.

    Was it necessary for the weight loss distribution to be normal? I answered no.
    no since clt applies
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    (Original post by xyzmaster)
    Part c used two independent samples. This part used only one sample that's why the wanted us to prove it
    Part D wants you to use U bar as mean, so that P(U1 > Ubar + S.D) = P(U1 > mean + SD), then do the normal distribution and you get P(z>1) as the means and S.Ds cancel out each other, I believe. I'd think doing that is worth 6 marks.
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    (Original post by simonli2575)
    A bit tricky, but generally I was happy when I walked out of the room.

    Was it necessary for the weight loss distribution to be normal? I answered no.
    No we need not assume it is normal because CLT does it for us
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    way too *****y this paper.... How the tell are we suppose to know all this ****...
    Edexcel is the best at "changing lives"...


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    (Original post by llIllIllIllI)
    Part D wants you to use U bar as mean, so that P(U1 > Ubar + S.D) = P(U1 > mean + SD), then do the normal distribution and you get P(z>1) as the means and S.Ds cancel out each other, I believe. I'd think doing that is worth 6 marks.
    Yessss did the same
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    (Original post by c3l6founder)
    no since clt applies
    (Original post by Navo D.)
    No we need not assume it is normal because CLT does it for us
    Lol my friends told me it was necessary because we needed to know it was normal before applying the CLT.
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    (Original post by Navo D.)
    Yessss did the same
    and then you get 0.159 to 3 d.p, which also makes sense
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    (Original post by xyzmaster)
    Pretty sure
    This is wrong unfortunately. They wouldn't give you 6 marks to repeat an answer. I went for P(z>sqrt(5)) = 0.0125 = 0.013 to 3dp... I'm most likely wrong but i thought I'd give it a shot. Its definitely NOT 0.181 to 3dp.
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    The question about whether or not you have to assume normal distributions what did people put for 2 marks? I put no you don't have to assume normal as we only need the mean of the sample to be normal using CLT
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    (Original post by llIllIllIllI)
    and then you get 0.159 to 3 d.p, which also makes sense
    1587 iirc yeah, hopefully we're both right. What'd you write for the last part of the stratified sampling question?
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    (Original post by Navo D.)
    Yessss did the same
    Isn't variance of ubar sigma^2/n n being 5 in this case
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    (Original post by MentalMath)
    The question about whether or not you have to assume normal distributions what did people put for 2 marks? I put no you don't have to assume normal as we only need the mean of the sample to be normal using CLT
    that's what I put, made it a bit longer to seemingly worth 2 marks...
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    So. Horribly. Wrong.
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    (Original post by xyzmaster)
    Isn't variance of ubar sigma^2/n n being 5 in this case
    Doesn't that give you the same ans? Since Ubar is a statistic of the same distribution it should just be Ubar and not have variance, or at least thats what I did
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    That one puzzled me too, I put no because it didn't ask for mean weight which would be CLT.
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    (Original post by Navo D.)
    1587 iirc yeah, hopefully we're both right. What'd you write for the last part of the stratified sampling question?
    I forgot what's it about already
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    Horrible paper, someone please do the unofficial!!!
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    (Original post by Navo D.)
    Doesn't that give you the same ans? Since Ubar is a statistic of the same distribution it should just be Ubar and not have variance, or at least thats what I did
    It gives p to be 0.181 and z value 0.91. Well we have proved have tricky this paper was that no one seems to be on agreement with this specific question
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    (Original post by Navo D.)
    Doesn't that give you the same ans? Since Ubar is a statistic of the same distribution it should just be Ubar and not have variance, or at least thats what I did
    I don't think the distribution for Ubar is required at all for that one
 
 
 
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