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S1 June 2016 UK Model Answers (Edexcel)

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Original post by kprime2
Yes



I don't have the paper yet


Alright Thanks anyway :smile:
Reply 81
Avatar for fowlerj
fowlerj
Original post by Rafeerahman
in the skewness question, did anyone write that the mean (3.43) was close to the median ( 3.47) so normal distribution can be used. i wanted to do the 3 (mean-median)/sd calculation but later i didnt do that uggghhhhh. i also wrote that the standard deviation he used (0.65) was a bit deviated from the standard deviation we had to find (0.68). i said despite these small inaccuracies, a normal distribution was fairly appropriate. am i getting 0 out of 3 for this?


I reckon you'll be able to say that they are close, so a normal distribution is appropriate. That's usually only 2 marks though. What worries me is that I didn't mention anything about the probability, P(W<3), which I think we were meant to include. I don't know what to say about that, though.

The actual data gives P(W<3)=0.18, which isn't very well supported by the model. (0.2546).
Original post by fowlerj
I reckon you'll be able to say that they are close, so a normal distribution is appropriate. That's usually only 2 marks though. What worries me is that I didn't mention anything about the probability, P(W<3), which I think we were meant to include. I don't know what to say about that, though.

The actual data gives P(W<3)=0.18, which isn't very well supported by the model. (0.2546).


They did they said use your answer to (e) and (f) or whatever it was , one of the parts it asked you to include was the question where you were asked find P(W<3)
Reply 83
Avatar for kprime2
kprime2
OP
20
Original post by HaveFa1th
i don't suppose you would get any credit for saying "little to no skew" so normal distribution is a good decision and data measured is continous?

Original post by Rafeerahman
in the skewness question, did anyone write that the mean (3.43) was close to the median ( 3.47) so normal distribution can be used. i wanted to do the 3 (mean-median)/sd calculation but later i didnt do that uggghhhhh. i also wrote that the standard deviation he used (0.65) was a bit deviated from the standard deviation we had to find (0.68). i said despite these small inaccuracies, a normal distribution was fairly appropriate. am i getting 0 out of 3 for this?



Original post by fowlerj
I reckon you'll be able to say that they are close, so a normal distribution is appropriate. That's usually only 2 marks though. What worries me is that I didn't mention anything about the probability, P(W<3), which I think we were meant to include. I don't know what to say about that, though.

The actual data gives P(W<3)=0.18, which isn't very well supported by the model. (0.2546).


5(e) I didn't realise it said use answer to part d too. In part d P(W<3) = 0.2546 whereas the actual P(W<3)=0.18. My comment would be that this just further consolidates that the normal model is not really accurate. Also, I don't think saying the decision to use the normal model on the basis that the median and median are close will be accepted. While the median and mean are close, the data is not nearly symmetrical, if it were part d P(W<3) would have yielded a more accurate approximation.
(edited 7 years ago)
Guys, for the first question, I got the right answer: 5.69 but then I rounded saying it was 5 (Whole wiggles). Reckon I'd lose a mark or something?

Also, what do you reckon the grade boundaries will be?
Reply 85
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SANTR
8
Original post by kprime2
Wouldnt be surprised


Really, can it fall as low as 55?
Reply 86
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kprime2
OP
20
Original post by SANTR
Really, can it fall as low as 55?


Well I thought this was quite a tricky paper. But it seems that the students found it easier than I think, so I think the boundary would be 56-57 for A.
Reply 87
Avatar for SANTR
SANTR
8
Original post by kprime2
Well I thought this was quite a tricky paper. But it seems that the students found it easier than I think, so I think the boundary would be 56-57 for A.


What's your predictions for C1?
Original post by SANTR
What's your predictions for C1?


I would say about 58-59
Reply 89
Avatar for PMONY
PMONY
1
Original post by kprime2
Definitely


Hi, I got the following:
C1: 67 marks
C2: 68 marks
S1: 56 marks

Would that be an A overall?
Reply 90
Avatar for mahmzo
mahmzo
17
Original post by PMONY
Hi, I got the following:
C1: 67 marks
C2: 68 marks
S1: 56 marks

Would that be an A overall?


Yes it is
Reply 91
Avatar for kprime2
kprime2
OP
20
Original post by kprime2
Well I thought this was quite a tricky paper. But it seems that the students found it easier than I think, so I think the boundary would be 56-57 for A.


Confirmed - 56 for an A. :u:

http://qualifications.pearson.com/content/dam/pdf/Support/Grade-boundaries/A-level/1606-AS-and-level-Grade-Boundaries-Final.pdf
so what would 90 UMS be bro and what would 85 UMS be too
Original post by kprime2




so what would 90 UMS be bro and what would 85 UMS be too
Original post by ALPACINO357
so what would 90 UMS be bro and what would 85 UMS be too


85 UMS is around 60/75 I think. 90 is about 63/75.
Original post by kprime2
Soz for the delay. Got caught up with the Euro games:colondollar:

I thought this was quite tricky
https://drive.google.com/open?id=0B3xSC6XqSy7kMGhaRHh2MHlnaDQ

EDIT: 5(e) I didn't realise it said use answer to part d too. In part d P(W<3) = 0.2546 whereas the actual P(W<3)=0.18. My comment would be that this just further consolidates that the normal model is not really accurate. Also, I don't think saying the decision to use the normal model on the basis that the median and median are close will be accepted. While the median and mean are close, the data is not nearly symmetrical, if it were part d P(W<3) would have yielded a more accurate approximation.



Hello,
Thank you for this and would you be able to possibly explain the working out for question 2e.

Thank you!
Reply 96
Avatar for kprime2
kprime2
OP
20
Original post by Rubydiamond3131
Hello,
Thank you for this and would you be able to possibly explain the working out for question 2e.

Thank you!


R = 1/X

the probabilities remain the same so multiply each new R by each probability
Original post by kprime2
R = 1/X

the probabilities remain the same so multiply each new R by each probability


Sorry - I don't understand where the -1/2 is from..
I am very sorry to bother you, I just needed some help!

Did you multiply the x value with the X value and then do 1 divided by this...
(edited 6 years ago)
Reply 98
Avatar for kprime2
kprime2
OP
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Original post by Rubydiamond3131
Sorry - I don't understand where the -1/2 is from..
I am very sorry to bother you, I just needed some help!

Did you multiply the x value with the X value and then do 1 divided by this...


1/(-2) = -1/2

R values multiplied by probabilities
Reply 99
Avatar for mahmzo
mahmzo
17
Original post by kprime2
1/(-2) = -1/2

R values multiplied by probabilities


Will you be doing model answers for C3 C4 and M1 this year :biggrin:

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