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Edexcel s1 June 2016 unofficial mark scheme

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Avatar for jbaten
jbaten
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How did you get a pic of the question and do u have the others if you could show all the others help would be much appreciated?
Original post by Mr T Pities You
Not the middle bit, sorry.
image.jpg
does the mean stay the same on that question about adding more numbers? and how many marks was that question
Original post by target21859
It seems I keep losing marks... :frown: First I thought I had 71 now it's more like 64. Probably less cause I left my my values for PMCC and standard deviation to 2dp (for some really annoying reason) and I'm not really confident I'll get the marks...

What are you supposed to leave your answers to? 3sf? I wouldn't worry about it I'm sure you got marks for working out for questions you got wrong
https://www.change.org/p/edexcel-as-level-maths-2016-c1-c2-s1-edexcel/c

Petition to lower grade boundaries.
Original post by neyjunior
for part e can you mention that the data was very slightly negatlively skewed and median and mean were close together therefore it can be modelled by a normal?


Yes. That's always been a valid answer that they allow. You can go either way with this question.
for the hotel question about people who had dinner i got 0, assuming that you had to use the probabilities given/ worked out in the first part of the question doesn't seem right to me because they gave new probabilities P(RnB) in the first part of the question was different from the new 37/77 probability. and for the last question, i did the probabilities of the intersection divided by 0.5 where the intersection was .18 *.5 (not sure if this is correct) because the intersection of the two probabilities is their product?
for the sarah and rebecca question i got like 0.016 or something like that, cant remember. for sarah winning, i did the probability of sarah getting her two highest scores (1.5, 2) added the probabilities these two and multiplied by the sum of rebecca's lowest( except for them getting negative 1 and a half). can someone reply please
Original post by veldt127
Ok, I managed to get all my answers stored in my calculator at the end, this is what I got. Not 100% sure on all of them, but we can compare.

Question 1

Spoiler

Question 2

Spoiler

Question 3

Spoiler

Question 4

Spoiler

Question 5

Spoiler

Question 6

Spoiler




What was the workings to questions 5e thanks
Original post by Katieramsay
How did everyone get the probability for Sarah and Rebecca wining. I got 0.4625 and 0.4450 so clearly I've got it wrong but I don't understand how?? 🤓🤓


I got that too, why is that wrong though?
Original post by Katieramsay
How did everyone get the probability for Sarah and Rebecca wining. I got 0.4625 and 0.4450 so clearly I've got it wrong but I don't understand how?? 🤓🤓

I did exactly the same thing we you. Do you think we'd get any marks? 😬
Original post by sharmonraja
I got that too, why is that wrong though?


Original post by Bret123
I did exactly the same thing we you. Do you think we'd get any marks? 😬


Maybe one mark? I hope we do either way get something!
Original post by Mr T Pities You
Check back through the thread for a pic. I know what you did wrong: the question said that a computer selected a single value of X, then both Sarah and Rebecca are allocated a score based on this using X or 1/X.
You treated it as one value of X picked for sarah, then a different value picked for Rebecca. A lot of calculation and cases later you got your answers.
You might get marks for a special case misread, as your question is actually harder but testing the same maths.


I got something like 0.4625 and 0.4450 as well, why is that wrong?
And what did you have to do to get the right answer?
Also how likely is it that I'll get 'special case' marks
Original post by studentsixth
Maybe one mark? I hope we do either way get something!


Ahh one mark out of four is disappointing, especially after how long i spent on that question
Original post by studentsixth
Maybe one mark? I hope we do either way get something!


Ahh hopefully special case marks or something 🙏🏼 For the PMCC did you say the effect of the values mean that it would decrease therefore more negative correlation
Original post by sharmonraja
Ahh one mark out of four is disappointing, especially after how long i spent on that question


I'm not sure to be honest! But there would only be two method marks and two answer marks. If we didn't get the answers right then we can only get a max of 2 :frown:
Original post by studentsixth
I'm not sure to be honest! But there would only be two method marks and two answer marks. If we didn't get the answers right then we can only get a max of 2 :frown:


Yeah I would be happy with 2 marks at least, but technically we still didnt get the correct method to get working out marks
Sooo.. we shall hope we can get something
Original post by PhilipBlanksky
What are you supposed to leave your answers to? 3sf? I wouldn't worry about it I'm sure you got marks for working out for questions you got wrong

Well the standard is 3sf but they didn't ask for how many they wanted so they shouldn't be able to penalise me for it hopefully.
Any idea what the grade boundaries might be? Found the probability section rather difficult
Original post by jbman690
for the hotel question about people who had dinner i got 0, assuming that you had to use the probabilities given/ worked out in the first part of the question doesn't seem right to me because they gave new probabilities P(RnB) in the first part of the question was different from the new 37/77 probability. and for the last question, i did the probabilities of the intersection divided by 0.5 where the intersection was .18 *.5 (not sure if this is correct) because the intersection of the two probabilities is their product?


Ok basically, you need to find P(D intersect B) and P(D intersect R {not b}) so you can use the probabilities in first part because it said estimate. So I just did a ratio as P(B) = 0.33 +0.27 = 0.6 and P(B intersect D) = 0.27 so I did (0.27/0.6)*40 = 18 people.
Then you knew that just P(R not B) = 0.22+0.15 = 0.37 and P(R {not B} intersect D) = 0.15 so i did (0.15/0.37)*37 = 15 people
So the total people who had dinner were 18+15 = 33
so P(Dinner) = 33/77 = 3/7
Screen Shot 2016-06-15 at 19.30.07.png43.9KB
2016-06-15.jpg95.2KB
Original post by sharmonraja
I got something like 0.4625 and 0.4450 as well, why is that wrong?
And what did you have to do to get the right answer?
Also how likely is it that I'll get 'special case' marks


See post 135

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