# OCR (non mei) S2 Wednesday 15th June 2016

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

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/...papers/s2-ocr/

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/...papers/s2-ocr/

0

reply

Report

#3

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.

0

reply

Report

#4

(Original post by

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.

**hallo.C**)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.

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? ??? ? ?

0

reply

Report

#5

(Original post by

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? ??? ? ?

**duncanjgraham**)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? ??? ? ?

0

reply

Report

#6

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?

0

reply

Report

#7

(Original post by

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?

**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?

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.. . . .

0

reply

Report

#8

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!

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!

0

reply

(Original post by

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!

**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!

Hope this helps

0

reply

Report

#10

(Original post by

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

**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

0

reply

(Original post by

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.

**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.

0

reply

Report

#12

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.

0

reply

Report

#13

(Original post by

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.

**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.

0

reply

Report

#14

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

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

0

reply

Report

#15

(Original post by

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.

**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.

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%20S...on%20Sheet.pdf to be helpful. Basically, do it when you are approximating a discrete distribution using the CLT.

0

reply

Report

#16

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!

0

reply

Report

#17

(Original post by

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%20S...on%20Sheet.pdf to be helpful. Basically, do it when you are approximating a discrete distribution using the CLT.

**PotAuFeu**)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%20S...on%20Sheet.pdf to be helpful. Basically, do it when you are approximating a discrete distribution using the CLT.

0

reply

Report

#18

(Original post by

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!

**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!

0

reply

Report

#19

**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!

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 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.

0

reply

X

### Quick Reply

Back

to top

to top