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AQA M1SB 8th of june 2016 unofficial markscheme watch

• View Poll Results: [AQA MS1B 8th of June 2016] What raw mark do you think the A grade will be ?
66
20.00%
65
5.00%
64
10.00%
63
8.33%
62
13.33%
61
11.67%
60
15.00%
59
5.00%
58
11.67%

1. Do we need above 95% in each maths module to study maths in oxford???
2. (Original post by goofyygoober)
Do we need above 95% in each maths module to study maths in oxford???
When i've been to their courses they have not said that they want above 95% ums, as in it being an entry requirement but usually getting the higher end of the marks is favourable, especially amongst the competition.
3. (Original post by JamieOH)
My answers (no idea if they are right ir not just contributing):

1.)a.) r=0.9591...
b.) Strong correlation blah blah

2.)a.)i.) 26
ii.) Median: 25, IQR: 2
b.) Mean and range
c.) Mean: 25.6, Range: 45

3.)a.)i.) 0.352
ii.) 0.136
iii.) 0.158
iv.) 0.792
v.) 0.267
b.) 0.00513

4.)a.) y= 181.3 + 3.0042x
b.) (context of gradient)
c.) 385
d.) I put the estimate was accurate as even though the residuals are large so are the values of Y but this could well be wrong.

5.)a.)i.) 0.941
ii.) 0.149
iii.) 0.792
iv.) Unity
b.) 1522.7
c.)i.) 0.355
ii.) 0.993

6.)a.)i.) 0.0262
ii.) 0.880
iii.) 0.101
iv.) 0.891
b.) Mean: 54, Var: 44.28

7.)a.) (257.92,377.08)
b.)i.) 381 not equal to 400 so not valid
ii.) 7/40 not equal to 0.25 so not valid (which is wrong, it IS valid but i didnt read the question properly)

Hope this helps, please point out if I went wrong somewhere which I'm sure i did in multiple places.
how did you get 0.00513 in question 3b? I think I did the same but why others get 0.0308?
4. (Original post by goofyygoober)
how did you get 0.00513 in question 3b? I think I did the same but why others get 0.0308?
Yes, I think we were supposed to multiply by 6 (3 combinations, 3!). We were wrong.
5. For question 3b I thought, to account for different combination you had to multiply by 4! (24) because there where four conditions to be met, so the amount of times they could be ordered is 4!? or are A(1) A(2) B(1) B(2) and A(2) A(1) B(1) B(2) considered to be the same and so you don't have to include both their probabilities, etc.?
6. (Original post by Dapperblook22)
it wasn't the first 10, as otherwise you would be comparing pounds to euros. You had to either convert the euros to pounds, or the pounds to euros and then find that it was 7/40 entries.
Im sure the table were all in euros and the answer had to be in euros? There was no mention of pounds... So no conversion required? Did i read the question wrong
7. (Original post by Shaks.smxth)
Im sure the table were all in euros and the answer had to be in euros? There was no mention of pounds... So no conversion required? Did i read the question wrong
I think you may have misread it :/ the table was definitely in £ and it told you the conversion was £1 to 1.2 euros. It was trying to see whether you would multiply the mean and SD they gave you when they wanted answers in Euros.
8. (Original post by Cascadess)
For question 3b I thought, to account for different combination you had to multiply by 4! (24) because there where four conditions to be met, so the amount of times they could be ordered is 4!? or are A(1) A(2) B(1) B(2) and A(2) A(1) B(1) B(2) considered to be the same and so you don't have to include both their probabilities, etc.?
It's 4!/2!*2!, 2!*2! because there are two unique events and they both occur twice.

I had to verify it in the exam by drawing a probability tree to check I had all the combinations covered.
9. (Original post by Cascadess)
I think you may have misread it :/ the table was definitely in £ and it told you the conversion was £1 to 1.2 euros. It was trying to see whether you would multiply the mean and SD they gave you when they wanted answers in Euros.
Oh crap! REALLY
Im such an idiot urgh. Thanks for the correction though You probably got it right then!
10. (Original post by Shaks.smxth)
Oh crap! REALLY
Im such an idiot urgh. Thanks for the correction though You probably got it right then!
No Problem! Sorry to break that news to you :/ And It was a bit of a fiddly question, so I'm not sure I got it all right, but thanks!
11. (Original post by -jordan-)
It's 4!/2!*2!, 2!*2! because there are two unique events and they both occur twice.I had to verify it in the exam by drawing a probability tree to check I had all the combinations covered.
uuuggghhhhh
12. Q7. I multiplied by CI by 1.2 to get my CI in Euros, was roughly (309,450 or something), I then said that the statement was invalid as it was possible for the mean to be below 400, due to the 309 being the lower bound of the CI? How many marks, if any, do you think I wouldn't got for that?
13. (Original post by -jordan-)
It's 4!/2!*2!, 2!*2! because there are two unique events and they both occur twice.

I had to verify it in the exam by drawing a probability tree to check I had all the combinations covered.
Okay i didnt do that... I did (condition A)^2 + (condition B)^2. Then i subtracted this from 1... Because these four people were not part of the population of 500.

Now I realise that I shouldn't have added the two probablilities
Does that make sense or is that completely wrong?
14. (Original post by Shaks.smxth)
Okay i didnt do that... I did (condition A)^2 + (condition B)^2. Then i subtracted this from 1... Because these four people were not part of the population of 500.

Now I realise that I shouldn't have added the two probablilities
Does that make sense or is that completely wrong?
If I am understanding what I DID wrong, then I think you should have multiplied where you added because you are finding the probability of independent event occurring, and then multiplied all of that by 6 to account for the different orders the people could be picked out.
15. (Original post by Cascadess)
If I am understanding what I DID wrong, then I think you should have multiplied where you added because you are finding the probability of independent event occurring, and then multiplied all of that by 6 to account for the different orders the people could be picked out.
I'm not sure if you were supposed to square the probability because it wasn't from the sample they gave you or it was dependent and you had to decrease the denominator. I did 6* 164/500*163/499*121/498*120/497 or something similar, can't remember the exact numbers but now I'm doubting whether that was correct.
16. (Original post by xs4)
i thought it was the first 10 that bought less than 200
=> 10/40 = 25%
The values on the table were the amounts they paid in pounds. It asked if less than 25% of people bought fewer than 200 euros. This means only the first 7 would have bought fewer than 200 euros because you had to times it by 1.2.

7/40 x 100 = 17.5%

At least that's what I put, not sure.
17. (Original post by -jordan-)
I'm not sure if you were supposed to square the probability because it wasn't from the sample they gave you or it was dependent and you had to decrease the denominator. I did 6* 164/500*163/499*121/498*120/497 or something similar, can't remember the exact numbers but now I'm doubting whether that was correct.
Because it was not from their sample, there was no decreasing denominators I'm pretty sure. You where just using the results in the table on an infinite population of people at the dentists.

So I did 164/500 * 164/500 * 121/500 * 121/500 * 4!

Which is the same as (164/500)^2 * (121/500)^2 * 4!

Because we are not taking our selecting from our sample, I don't think the sample size goes down.

(Btw i have no idea if those numbers are correct, i'm just going off you here )
18. (Original post by Sam Webb)
The values on the table were the amounts they paid in pounds. It asked if less than 25% of people bought fewer than 200 euros. This means only the first 7 would have bought fewer than 200 euros because you had to times it by 1.2.

7/40 x 100 = 17.5%

At least that's what I put, not sure.
It was 8/40 because one of them was equal to 200 euros.
19. (Original post by JamieOH)
My answers (no idea if they are right ir not just contributing):

1.)a.) r=0.9591...
b.) Strong correlation blah blah

2.)a.)i.) 26
ii.) Median: 25, IQR: 2
b.) Mean and range
c.) Mean: 25.6, Range: 45

3.)a.)i.) 0.352
ii.) 0.136
iii.) 0.158
iv.) 0.792
v.) 0.267
b.) 0.00513

4.)a.) y= 181.3 + 3.0042x
b.) (context of gradient)
c.) 385
d.) I put the estimate was accurate as even though the residuals are large so are the values of Y but this could well be wrong.

5.)a.)i.) 0.941
ii.) 0.149
iii.) 0.792
iv.) Unity
b.) 1522.7
c.)i.) 0.355
ii.) 0.993

6.)a.)i.) 0.0262
ii.) 0.880
iii.) 0.101
iv.) 0.891
b.) Mean: 54, Var: 44.28

7.)a.) (257.92,377.08)
b.)i.) 381 not equal to 400 so not valid
ii.) 7/40 not equal to 0.25 so not valid (which is wrong, it IS valid but i didnt read the question properly)

Hope this helps, please point out if I went wrong somewhere which I'm sure i did in multiple places.
For "7)b)i)" The confidence interval from the previous question is in Pounds. So I think you have to multiply both limits by 1.2 to give the confidence interval in Euros. Which makes it so that 400 does lie within the confidence interval.
20. (Original post by Sam Webb)
The values on the table were the amounts they paid in pounds. It asked if less than 25% of people bought fewer than 200 euros. This means only the first 7 would have bought fewer than 200 euros because you had to times it by 1.2.

7/40 x 100 = 17.5%

At least that's what I put, not sure.
this is what i put too, i messed up the CI tho, put the wrong values in, does this mean i loose all 4 marks? and will i get ECF for the claim, as normally markschemes say M1 for comapring the CI with the given value? so should only loose 3 marks max here??

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