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# The Official Edexcel S2 Jan 2010 Thread watch

1. I cant wait, havent done much revision, and the exam is in about 8 hours, i need to sleep, but watch me ace this thing

anyone else who cant sleep? :P
2. Locking the thread while it's quiet until after the time limit at 12.00am tonight, when you can discuss the post exam
3. So, how did everyone find it?
What do you think grade boundaries will be like?
4. I found it quite easy apart from the last subquestion which I didnt revise. Just waiting for answers to be posted!
5. (Original post by amzmalhotra)
I found it quite easy apart from the last subquestion which I didnt revise. Just waiting for answers to be posted!

I luckily revised that this morning
6. What was the last subquestion?
7. Anyone has answers to the test? Did any of you ask the teacher for them? I will rep for answers
8. From the ones I can remember:

The one that turns out to be a continuous uniform distribution, part (e) P(X=1) = 0

Last question. You find the mean the usual way and the variance for that matter, list out all possible samples, there are 8.

Probability of mean value
1 is 1/64,
4/3 is 9/64
5/3 is 27/64
2 is 27/64

You had to use Po(1/4) and Po(2/5) For the robot.

Probability that it breaks down in the subsequent 8 hours is 2/5, mutually exclusive.

Really cannot remember much more, if anyone can jog my memory, I can try and verify.

question 1.
a) A~B(20,0.05)
b) 0.3585
c) 0.0026
d) mean=1 Var(x)=0.95

question 2.
a) 1/12
b) f(x)=1/6 for -2<=x<=4
f(x)=0 otherwise

c) Uniform rectangular
d) Var(x)=3
e) 0

question 3.
a) 0.7788
b) 0.3297
c) 0.0536
d) 0.03297 (independence)

question 4.
a) k=1/9
b) F(x)=0 for x<=0
F(x)=1/9( x^3/3 -x^2 +2x) for 0<x<=3
F(x)=x/3 -1 for 3<x<=4
F(x)=1 for x>=4

c) mean=2.4167
d) I only used the equation for 0-3 range, change of sign indicates root in between

question 5.
a) 0.3328
b) 0.0485

question 6.
a) x<=2 x>=16
c) 0.85%

question 7.
a) mean=1.75 Var(x)=0.1875
b) list of all samples
c) increasing mean, prob: 1/64, 9/64, 27/64, 27/64

DISCLAIMER: THEY ARE NOT 100% RIGHT ANSWERS, THEY ARE JUST MY ANSWERS FROM THE TEST, THERE ARE MISTAKES HERE.
10. Oh yh, last question was on sampling dist.
11. (Original post by Clarity Incognito)
You had to use Po(1/4) and Po(2/5) For the robot.

Probability that it breaks down in the subsequent 8 hours is 2/5, mutually exclusive.
Damn got that wrong then, I put that it would be the answer you worked out for p(x>=1) because breakdowns are indpendant/happen at a constant rate.
12. (Original post by Moa)

question 1.
a) A~B(20,0.05)
b) 0.3585
c) 0.0026
d) mean=1 Var(x)=0.95

question 2.
a) 1/12
b) f(x)=1/6 for -2<=x<=4
f(x)=0 otherwise

c) Uniform rectangular
d) Var(x)=3
e) 0

question 3.
a) 0.7788
b) 0.3297
c) 0.0536
d) 0.03297 (independence)

question 4.
a) k=1/9
b) F(x)=0 for x<=0
F(x)=1/9( x^3/3 -x^2 +2x) for 0<x<=3
F(x)=x/3 -1 for 3<x<=4

was it not x/3 -1/3

F(x)=1 for x>=4

c) mean=2.4167
d) I only used the equation for 0-3 range, change of sign indicates root in between

question 5.
a) 0.3328
b) 0.0485

question 6.
a) x<=2 x>=16
c) 0.85%

hypoth. testing question right, 15 was not in critcial region. actual significan was like 0.7%, summin less than 1%

question 7.
a) mean=1.75 Var(x)=0.1875
b) list of all samples
c) increasing mean, prob: 1/64, 9/64, 27/64, 27/64
was it not x/3 -1/3

hypoth. testing question right, 15 was not in critcial region. actual significan was like 0.7%, summin less than 1%
13. (Original post by Moa)

question 1.
a) A~B(20,0.05)
b) 0.3585
c) 0.0026
d) mean=1 Var(x)=0.95

question 2.
a) 1/12
b) f(x)=1/6 for -2<=x<=4
f(x)=0 otherwise

c) Uniform rectangular
d) Var(x)=3
e) 0

question 3.
a) 0.7788
b) 0.3297
c) 0.0536
d) 0.03297 (independence)

question 4.
a) k=1/9
b) F(x)=0 for x<=0
F(x)=1/9( x^3/3 -x^2 +2x) for 0<x<=3
F(x)=x/3 -1 for 3<x<=4
F(x)=1 for x>=4

c) mean=2.4167
d) I only used the equation for 0-3 range, change of sign indicates root in between

question 5.
a) 0.3328
b) 0.0485

question 6.
a) x<=2 x>=16
c) 0.85%

question 7.
a) mean=1.75 Var(x)=0.1875
b) list of all samples
c) increasing mean, prob: 1/64, 9/64, 27/64, 27/64
Good memory, I agree with all of q.1, 2 and 7. Unfortunately I can't remember enough of the rest to be sure. Anwyays, hi5.

EDIT Apart from 2a) which I've just read again, and other people have checked.
14. (Original post by Moa)

question 1.
a) A~B(20,0.05)
b) 0.3585
c) 0.0026
d) mean=1 Var(x)=0.95

question 2.
a) 1/12
b) f(x)=1/6 for -2<=x<=4
f(x)=0 otherwise

c) Uniform rectangular
d) Var(x)=3
e) 0

question 3.
a) 0.7788
b) 0.3297
c) 0.0536
d) 0.03297 (independence)

question 4.
a) k=1/9
b) F(x)=0 for x<=0
F(x)=1/9( x^3/3 -x^2 +2x) for 0<x<=3
F(x)=x/3 -1 for 3<x<=4
F(x)=1 for x>=4

c) mean=2.4167
d) I only used the equation for 0-3 range, change of sign indicates root in between

question 5.
a) 0.3328
b) 0.0485

question 6.
a) x<=2 x>=16
c) 0.85%

question 7.
a) mean=1.75 Var(x)=0.1875
b) list of all samples
c) increasing mean, prob: 1/64, 9/64, 27/64, 27/64

I got mostly the same answers but for the cdf i got 1/3x -1/3
15. Also someone said high grade boundaries. How high approx. 68+ for an A?
16. Posting the whole paper in few seconds
17. I thought that was a very good paper. I expect boundaries to at least 65/66
18. qstns 2(a), was mean not 1, 1/2(a+b), can see this from f(x) in later part of question
19. Edexcel S2 - Tuesday 19 January 2010 Paper: (warning, very poor quality)
Spoiler:
Show

20. (Original post by Moa)

question 1.
a) A~B(20,0.05)
b) 0.3585
c) 0.0026
d) mean=1 Var(x)=0.95

question 2.
a) 1/12
b) f(x)=1/6 for -2<=x<=4
f(x)=0 otherwise

c) Uniform rectangular
d) Var(x)=3
e) 0

question 3.
a) 0.7788
b) 0.3297
c) 0.0536
d) 0.03297 (independence)

question 4.
a) k=1/9
b) F(x)=0 for x<=0
F(x)=1/9( x^3/3 -x^2 +2x) for 0<x<=3
F(x)=x/3 -1 for 3<x<=4
F(x)=1 for x>=4

c) mean=2.4167
d) I only used the equation for 0-3 range, change of sign indicates root in between

question 5.
a) 0.3328
b) 0.0485

question 6.
a) x<=2 x>=16
c) 0.85%

question 7.
a) mean=1.75 Var(x)=0.1875
b) list of all samples
c) increasing mean, prob: 1/64, 9/64, 27/64, 27/64

I got 1/3 for 2a
i got 1/3x -1/3 for cdf
when you put in 2.6 and 2.7 in cdf, the value comes below 0.5 and above 0.5 and this shows it lies in the median

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