OCR (non mei) S2 Wednesday 15th June 2016

Hi!This is for discussion pre and post the up coming S2 exam :smile:

Lets start the discussion!
For past papers and mark schemes:
http://www.ocr.org.uk/qualifications/as-a-level-gce-mathematics-3890-3892-7890-7892/
For some older past papers:
http://www.physicsandmathstutor.com/a-level-maths-papers/s2-ocr/

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this is quiet

I am struggling to figure out when you divide the variance by the sample size and when you don't. I thought that you didn't divide by sample size when the original distribution is normal but I did a past paper question where you did.

Reply 3

duncanjgraham

Original post by hallo.C

generally, you should divide by n when there is a question involving a sample

You do divide by n when the original distribution X is normal , however you are not using the central limit theorem, because the distribution of the sample mean is exactly normal.

hope this helps? ??? ? ?

(edited 7 years ago)

Reply 4

hallo.C

Original post by duncanjgraham

So is it only if I have worked out an unbiased estimate of a distribution that is originally normal, that I don't divide the variance by sample size?

Reply 5

hallo.C

Sorry to keep pushing this but I don't understand. June 2008 question 3, you divide by sample size when it's normal and then on question 2 January 13, you don't divide by sample size because it's normal?

Reply 6

duncanjgraham

Original post by hallo.C

jan 13 is an absolute b i t c h of a paper, i just did it today.

q 2 is a bit of a special case, it's not asking for the distribution of the sample mean of C - (C with a line on top of it) - and is just interested in the distribution of C, and using the data (sum of x/ n etc) helps estimate the distribution of C.

if you were asked to find the distribution of the sample mean of C for this question, you would divide the variance by n.

Your query is quite hard to help you with, you may need a teacher.. . . .

(edited 7 years ago)

Reply 7

Nl1998

Hi,

You know for some of the normal distribution questions, in the mark scheme they always give the z value to 4 decimal places. But, the tables only give you an answer from 3 decimal places. So, I get a different answer sometimes from what they accept. Does that matter or will they still accept my answer?Thanks!

Reply 8

TheNicholas

Original post by Nl1998

Hey, in the mark scheme is normally says if it rounds to ... So say it says 0.6547 it will also say in the side bit or r.t 0.655,
Hope this helps :smile:

Reply 9

Nl1998

Original post by TheNicholas

Hey, in the mark scheme is normally says if it rounds to ... So say it says 0.6547 it will also say in the side bit or r.t 0.655,
Hope this helps :smile:

Yeah but sometimes because of calculating to a lower degree of accuracy, the answer is slightly different and still doesn't round to the the answer in the markscheme even to 3s.f.

Reply 10

TheNicholas

Original post by Nl1998

Yeah but sometimes because of calculating to a lower degree of accuracy, the answer is slightly different and still doesn't round to the the answer in the markscheme even to 3s.f.

Umm personally that's never happened to me and I've done all the papers, are you sure you're using continuity corrections or something like that???

Reply 11

PotAuFeu

Just had a look at the previous few years' grade boundaries and jeez are they high! I was feeling a bit confident about this but now I'm not so sure.

Reply 12

duncanjgraham

Original post by PotAuFeu

Just had a look at the previous few years' grade boundaries and jeez are they high! I was feeling a bit confident about this but now I'm not so sure.

very high indeed,

Reply 13

ishanS

Not looking forward to this at all
Can anyone help me with when to use the 1/2n continuity correction please? i have no idea!
Also in some of the mark schemes it says it will give you the full marks even if your continuity correction is incorrect/used wrong/ not even used so im relying on that to do well in stats lol

Reply 14

PotAuFeu

Original post by duncanjgraham

Fck it's hard to give examples but basically if you were correcting P(X<5) you would make it P(X<5+0.5) but if it's a case where 1/2n is required you actually make P(X<5) into P(X<5minus1/2n)

Likewise you'd make P(X>5) into P(X>5+1/2n) where you would normally make it P(X>5-0.5)

I hope this helps.

Are you sure that the continuity correction for P(X<5) is P(X<5+0.5)?
The way I remember it is that you want it <5 so take away 0.5 so you don't include 5 and if you want >5 then add 0.5 so you don't include 5. Likewise, if you want <=5 then add 0.5 to include 5 and if you want >=5 then you take away 0.5 to include 5.
Hope that makes sense, I haven't seen particularly many 1/2n corrections but when I have, I just do my continuity corrections as I said so above and the answer comes out fine.

For when to use 1/2n, I found http://www.mathshelper.co.uk/OCR%20S2%20Revision%20Sheet.pdf to be helpful. Basically, do it when you are approximating a discrete distribution using the CLT.

Reply 15

emmacurtis8

Can someone please explain to me when I do and when I do not divide the variance by n - it would be very much appreciated!

Reply 16

16characterlimit

Original post by PotAuFeu

P(X<5) is P(X<5 -0.5) as I remember it as well.

Reply 17

PotAuFeu

Original post by emmacurtis8

Can someone please explain to me when I do and when I do not divide the variance by n - it would be very much appreciated!

Someone will definitely explain this better than me but whenever you are working with the mean of n occurrences (a sample) you need to divide variance by n. In the question it may explicitly mention "the mean of n occurrences" or "a random sample of n things", or you might see x-bar (x with a line over the top) which indicates that you need to use the mean of the sample. Pretty much if it mentions mean (and you're using the normal distribution) then you need to divide the variance by n.

Reply 18

16characterlimit

Original post by emmacurtis8

Can someone please explain to me when I do and when I do not divide the variance by n - it would be very much appreciated!

I'm not absolutely 100% sure, but this is what I know so far and it's been working, it would be good for someone to double check me.

For a start if you ever have a discrete distribution (Binomial, Poisson) and approximate it with normal, you will NEVER (in S2 OCR) have to divide the variance by n, I don't know why, you just don't.

If you are told, the "population variance" or the question implies somehow that the parameters for your distribution are known for the entire population instead of either given or worked out from a sample, you DON'T divide by n. After looking at some papers, ignore this.

If you have used the central limit theorem, then divide by n (duh), also apply the 1/2n correction if the distribution/data was discrete.

If you see the distribution letter (X, S, Y... e.t.c) with a bar, divide by n.

This isn't as reliable as the ones above, but if you see the words "sample mean" you 70% are likely to divide by n.