S1 Jan 2011 Edexcel - Solutions (and paper) in the OP (Now On)

I would be amazed if it wasn't below 60 for 80UMS

I really hope your right! It was a hard paper defo but i dint think the gb would be lower thsn 60 but then agen i have no basis to judge the gb on, whereas you know what your tlking about:P

Reply 21

Arsey

OP

10

Original post by CapsLocke

your regression line seems to be wrong.

you done Svp/Svv, and they hadn't given you you Svp, so shouldn't you of done Spv/Spp to work out B?

in the standard regression formula

b = Sxy/Sxx

In this question v = (x - 5)/10 so v relates to x and p relates to y

so Sxy is Spv (or Svp same thing) and Sxx is Svv NOT Spp; so I you are wrong I am afraid.

Reply 22

perrytheplatypus

14

That paper seemed easier...well easier than the one with the diseased dog probablity one anyway.

It can't used to have been 69-75 for an A in all those papers?

Reply 23

Edwin Okli

3

Original post by CapsLocke

your regression line seems to be wrong.

you done Svp/Svv, and they hadn't given you you Svp, so shouldn't you of done Spv/Spp to work out B?

The amount of peas (p) depended on the amount of rainfall (v). Therefore peas were the y and rainfall was the x. They even tell you that the uncoded rainfall was x anyway so you should have thought about it that way. Plus, they told you to work out the regression line "p = a + bv", so again p = y and v = "x".

Oh and Spv = Svp.

Reply 24

CapsLocke

11

Original post by Arsey

in the standard regression formula

b = Sxy/Sxx

In this question v = (x - 5)/10 so v relates to x and p relates to y

so Sxy is Spv (or Svp same thing) and Sxx is Svv NOT Spp; so I you are wrong I am afraid.

o snap, that's tricky to spot.

Reply 25

shay_5293

1

Original post by perrytheplatypus

That paper seemed easier...well easier than the one with the diseased dog probablity one anyway.

It can't used to have been 69-75 for an A in all those papers?

Ur lookin at 100 ums. I think those gb show 100 90 80 70.... ums in that order, so the third one along is thw A

Reply 26

Arsey

OP

10

Original post by Mira1

Hi Arsey

Question 3c I got 41.8 for the median and for the skew, i thought it was -ve skew because I worked out the LQ as 32 and the UQ as 48.6 and showed working for them and the median. then i did Q3-Q2 < Q2-Q1 to give a negative skew. I got part b right. How many marks do you think ill lose on that question?

You wouldn't be expected to go through interpolation to work out Q1 and Q3 to find describe skew unless it was already asked. However, what you did seems to be correct. Looking at the shape of the data it does look like negative skew. Unfortunately it is because the actual data must be closer to the upper ends of the boundaries.

It is stated that Sum t = 1414 but if you work it out using midpoints it is 1314. I thought that was really strange to the extent I suspected an error may have been made in the question. I have never seen a question like it.

Reply 27

Arsey

OP

10

Original post by Edwin Okli

The amount of peas (p) depended on the amount of rainfall (v). Therefore peas were the y and rainfall was the x. They even tell you that the uncoded rainfall was x anyway so you should have thought about it that way. Plus, they told you to work out the regression line "p = a + bv", so again p = y and v = "x".

Oh and Spv = Svp.

and x was coded to give v.

I don't think it was difficult to realise that v related to x.

Reply 28

BeritV

9

I honestly hope that you are right. If you are, I got 75/75 :-)

Reply 29

Arsey

OP

10

Original post by perrytheplatypus

That paper seemed easier...well easier than the one with the diseased dog probablity one anyway.

It can't used to have been 69-75 for an A in all those papers?

being pedantic, you can not get an A in any individual module, just a UMS.

All the past boundaries are in the first post.

The majority of the time it was below 60 for 80UMS.

June 2010 was tricky but I thought this was more difficult than a standard S1 paper.

Reply 30

ALZA

I LOVE YOU ARSEY :biggrin:

Only thing I found really difficult was that part in question 6.. never seen it before in a past paper and it isn't in my book either, I meen wtf :s-smilie:

?? Would it be around 60 marks for an A?

Reply 31

beelz

Hey Arsey, i really need to know, please please please tell me there is error carried forward. on the tree diagram i completely messed up the values cos i didnt pay enough attention and that meant the whole of the question was ruined for me. i showed the correct working for all the questions except i obviously couuldnt do the ones which said show this equals this argghhh FML.

on the last question aswell i accidentally did 152 - u = s.d. instead of u - 152 = s.d. other than this i did all the working correctly but it gave me a different answer obviously, so how many marks out of 6 do you think ill get for this question? and how many for the tree diagram question?

Reply 32

Edwin Okli

3

Original post by Arsey

You wouldn't be expected to go through interpolation to work out Q1 and Q3 to find describe skew unless it was already asked. However, what you did seems to be correct. Looking at the shape of the data it does look like negative skew. Unfortunately it is because the actual data must be closer to the upper ends of the boundaries.

It is stated that Sum t = 1414 but if you work it out using midpoints it is 1314. I thought that was really strange to the extent I suspected an error may have been made in the question. I have never seen a question like it.

I seriously hope that was a mistake on the part of the exam board. They didn't even explain what t and t^2 meant and the mean and standard deviation question wasn't worth enough to require having to work out fx and fx^2 yourself. Of course, they could have calculated those data using the raw data and therefore using midpoints of course would vary.

Reply 33

beelz

also for question 5 i used n+1/2 for the median how many marks out of two will i get?

Reply 34

freakingout

i think i got q-8 mean correct (=155? that's what is given in the other thread) but i dont remember if i got the s.d correct or not, isnt it obvious that i got s.d correct? i got my mean using my s.d value
and can anyone say what was the lowest value for boxplot? wouldnt only 1 mark be cut if i gave two whiskars? :|

Reply 35

freakingout

Original post by Edwin Okli

I seriously hope that was a mistake on the part of the exam board. They didn't even explain what t and t^2 meant and the mean and standard deviation question wasn't worth enough to require having to work out fx and fx^2 yourself. Of course, they could have calculated those data using the raw data and therefore using midpoints of course would vary.

how did you calculate mean and s.d?
and whats the correct way?
:confused:

i used the given info t and t^2 :frown:

Reply 36

maths134

13

So calculating Q1 and Q3 to give a negative skewness is incorrect ? because when doing so it does actually give a negative skewness, why is that actually wrong ?

Reply 37

CapsLocke

11

last question I used the normal table instead of the percentiles, my answer wasn't far off you yours, I got 154 for the mean and 1.606 for the standard deviation.

I pretty much used the same method, how many marks do you think I would lose?

Reply 38

ghostkillah

6

Well I failed that. Why did both Jun 2010 and Jan 2011 have to be the hardest S1 papers I have done - I got A's in all the other ******* past papers

Reply 39

Arsey

OP

10

Original post by ALZA

I LOVE YOU ARSEY :biggrin:

Only thing I found really difficult was that part in question 6.. never seen it before in a past paper and it isn't in my book either, I meen wtf :s-smilie:

?? Would it be around 60 marks for an A?

question 6, do you mean the discrete random variables question? That was a pretty standard question, one of the easier ones on the paper I thought.

The new Edexcel Heinemann textbooks are very good (mostly).

S1 Jan 2011 Edexcel - Solutions (and paper) in the OP (Now On)

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