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# Edexcel S2 Exam Paper January 2011 14/01/11 Watch

1. (Original post by Red Devil k18)
ive seen the mark schem but dont understand how it works. if you add both prob's isnt it 1/6 +1/6=1/3 but answer is 2/3
no.........The first 1/6 comes from P(x>6). But if x<4 then the other side would be longer than 6cm as well. So P(x<4)=1/2. Add these probabilities together and you get 2/3
2. really silly question here - but I'm just checking... the exam is on the afternoon tomorrow not the morning... right??
3. (Original post by airey_head)
really silly question here - but I'm just checking... the exam is on the afternoon tomorrow not the morning... right??
Afternoon : 13:30pm
4. (Original post by LukethePianoMan)
I just thought what would be mean, asking to find critical regions after a normal approximation!
How would you do that? And are there any examples? Haven't seen this yet
5. (Original post by Ultimate1)
How would you do that? And are there any examples? Haven't seen this yet
I'm sure I haven't also seen any questions on those
6. (Original post by jit987)
Afternoon : 13:30pm
Haha Thankyouuu! Better to be safe than sorry :P

Good luck!
7. (Original post by jit987)
I'm sure I haven't also seen any questions on those
Yeah there have been a couple of questions like that on some of the earlier papers...
Just really really hope nothing like that comes up :/
8. (Original post by LukethePianoMan)
Nice to talk again
Pre exam thoughts? I am finding it pretty good so far, just hope nothing unusual comes up, like normal approximated critical regions or something!
Hello!

I'm finding it good, which feels like a shock to me! But what I am worried for is definitions >< Going to try and make sure that I am confident with them for tomorrow. Good luck!
9. (Original post by Ultimate1)
How would you do that? And are there any examples? Haven't seen this yet
I think you would look in the tables to find the z values of the appropriate significance level then code them back by multiplying by the standard deviation and adding the mean, then for the lower bound use the nearest integer larger than and for the upper bound use the nearest integer lower than, but this won't come up theres none in the book i dont think :s
10. I got two questions !! someone help me out before 5 hours !!

1.) A van company has 5 vans which they rent out by the day. Assuming that the number of vans hired out per day follows a Poisson distribution with mean 3, calculate, for a period of 100 days, the expected number of days when
-no vans will be hired
-the demand for the van is not satisfied

2.) The probability that a man is over 1.8m tall is 0.2 . A random sample of 50 men is taken and each man is measured and his height is recorded . Find the probability that the number of men in the sample over 1.8m tall is between 7 and 10 inclusive using :
-a Poisson approximation
-a normal approximation
I have found the answers to be as 0.4529(Poisson) and 0.4639(normal) and they are correct.
The next part of the question says 'By finding the actual value of the probability state which is the better approximation' . How do u go on solving this problem ? Help would be much appreciated !
11. (Original post by Ultimate1)
How would you do that? And are there any examples? Haven't seen this yet
Well if you think about it, you can do a hypothesis test on a normal distribution, like the last question on the S1 paper. I think you would make the null hypothesis based around big X (if that makes sense)

But chances of this happening, if it did it would fall into the critical region with a 1% sig level
Attached Images
12. S1 June 2010.pdf (277.7 KB, 114 views)
13. (Original post by cpdavis)
Well if you think about it, you can do a hypothesis test on a normal distribution, like the last question on the S1 paper. I think you would make the null hypothesis based around big X (if that makes sense)

But chances of this happening, if it did it would fall into the critical region with a 1% sig level
Thanks I kind of get it now. I don't think it will come up though as it is not in the new book of S2.
14. (Original post by warkan)
I got two questions !! someone help me out before 5 hours !!

1.) A van company has 5 vans which they rent out by the day. Assuming that the number of vans hired out per day follows a Poisson distribution with mean 3, calculate, for a period of 100 days, the expected number of days when
-no vans will be hired
-the demand for the van is not satisfied

2.) The probability that a man is over 1.8m tall is 0.2 . A random sample of 50 men is taken and each man is measured and his height is recorded . Find the probability that the number of men in the sample over 1.8m tall is between 7 and 10 inclusive using :
-a Poisson approximation
-a normal approximation
I have found the answers to be as 0.4529(Poisson) and 0.4639(normal) and they are correct.
The next part of the question says 'By finding the actual value of the probability state which is the better approximation' . How do u go on solving this problem ? Help would be much appreciated !
For part a, you would adjust the parameter by timings your mean by 100. To work out the number of days where you expect to have no vans hired, you do P(X=0) to get your probability and times it by the number of days. Not sure what the next part means

For the poisson and normal question, you would say that a normal approximation is better in this case. This is because p is not small and close to 0.5
15. (Original post by warkan)
I got two questions !! someone help me out before 5 hours !!

1.) A van company has 5 vans which they rent out by the day. Assuming that the number of vans hired out per day follows a Poisson distribution with mean 3, calculate, for a period of 100 days, the expected number of days when
-no vans will be hired
-the demand for the van is not satisfied
Got an answer for your first question... I'll do the other question once I've finished typing this!

X~ Po (3)
-For 1 day => P(X=0) = e^-3 = 0.4978...
For 100 days => Ans x 100 = 4.97...days => 5 days

-For 1 day => P(X > 5) = 1 - P(X<=5) = 1 - 0.9161 = 0.0839
For 100 days => Ans x 100 = 8.39 => 9 days

I hope this is actually right!!!
16. (Original post by warkan)

2.) The probability that a man is over 1.8m tall is 0.2 . A random sample of 50 men is taken and each man is measured and his height is recorded . Find the probability that the number of men in the sample over 1.8m tall is between 7 and 10 inclusive using :
-a Poisson approximation
-a normal approximation
I have found the answers to be as 0.4529(Poisson) and 0.4639(normal) and they are correct.
The next part of the question says 'By finding the actual value of the probability state which is the better approximation' . How do u go on solving this problem ? Help would be much appreciated !
I got the same answers as you for the first part of this question!

The better approximation is probably Normal as n is large and p is close to 0.5 i.e. for it to be Poisson, the value of p would have to be very small

But check mark schemes to see what the 'proper' answer is

And GOOD LUCK!!
17. Just did June 10 paper, found it relatively harder than the others. Would anyone else agree with this?
18. (Original post by airey_head)
Got an answer for your first question... I'll do the other question once I've finished typing this!

X~ Po (3)
-For 1 day => P(X=0) = e^-3 = 0.4978...
For 100 days => Ans x 100 = 4.97...days => 5 days

-For 1 day => P(X > 5) = 1 - P(X<=5) = 1 - 0.9161 = 0.0839
For 100 days => Ans x 100 = 8.39 => 9 days

I hope this is actually right!!!
Dude thanks so much !!! and btw the answers are correct!!! : )
19. Reading the spec, we only have to do hypothesis testing for poisson and binomial.

Also, when doing a critical region question, do about 3 c values for the > and < that way you show the examiner that you have explored that
20. (Original post by TSR-MATT)
Has anyone got the June 2009 S2 mark scheme, preferably in PDF.

I can only find it on word, and it won't open on my crappy version!
http://www.edexcel.com/migrationdocu...7_UA021531.pdf

They merged all of the marks schemes together that year
21. anyone else doind S1&S2 tomorow

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