# s2 OCR how did u find it??

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#41

Yea I proved it. It was something like P(x>50) then continuity correction meant it was P(X>=50.5) i cant find greater than or equals to sign. stupid continuity corrections!!! i couldnt quite remember how to do it in exam tho so i did it lots of different ways until i got the right answer!

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#42

(Original post by

yeh i proved that, i no all 3 types of approxs were iin the paper, wat other qs was hard, could everyone prve the long eqn, t4 + ............... etc

**starcrossed**)yeh i proved that, i no all 3 types of approxs were iin the paper, wat other qs was hard, could everyone prve the long eqn, t4 + ............... etc

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#43

(Original post by

I actually worked out the X values for which the regions would be rejected. I know that it was definitely two tailed. Thereofre you should have 2.5% at either end.

**Bhaal85**)I actually worked out the X values for which the regions would be rejected. I know that it was definitely two tailed. Thereofre you should have 2.5% at either end.

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#44

i had the 0.918 for the X~p(7) thought it was to big so i crossed it out and did it using P(X<=3)...good bye 3 marks

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#45

(Original post by

Yeah you had to integrate between t0 and 0 to equal 0.95. then times by the fraction, then by 4 and you should get the same.

**Bhaal85**)Yeah you had to integrate between t0 and 0 to equal 0.95. then times by the fraction, then by 4 and you should get the same.

i did all of that, except i didn't times by 4, lol i times by 250000 or something massive like that to get the same equation....

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#46

(Original post by

There was a hell of a lot of continutity(sp) corrections!!! anyone else prove that 0.231 thing?

**Bhaal85**)There was a hell of a lot of continutity(sp) corrections!!! anyone else prove that 0.231 thing?

Yeah, I managed to prove it in the end. First time I forgot the cc, the next time I used the wrong one, then I finally got it right. Wasted a lot of time on that question.

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#47

All together it seemed a pretty good paper, not like all the higher pure modules, there didnt seem to be anywhere you could bin a load of marks, maybe just a few small things here and there. What do you reckon the grade boundaries will be like? Because i need a high UMS mark in this to get the grades for my oxford offer

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#48

(Original post by

i did all of that, except i didn't times by 4, lol i times by 250000 or something massive like that to get the same equation....

**abc123**)i did all of that, except i didn't times by 4, lol i times by 250000 or something massive like that to get the same equation....

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#49

(Original post by

Thanks, that's what I did. I made a stupid mistake on question 2 and numbered the names 1-10, then got really confused. Maybe they'll got me a mark because they pity me.

**sock**)Thanks, that's what I did. I made a stupid mistake on question 2 and numbered the names 1-10, then got really confused. Maybe they'll got me a mark because they pity me.

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#50

(Original post by

You should have taken out the multiplied the 0.95 by 62500, then the entire equation by 4.

**Bhaal85**)You should have taken out the multiplied the 0.95 by 62500, then the entire equation by 4.

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#51

(Original post by

I labelled them from 1-10 too, but then realised that I should have done 0-9, so I changed it, lol. It threw me quite a bit that question, how did people answer it? As this is something we don't normally do. I mean, what the hell, tables of random numbers? People pay for this crap? How does it work? lol. I then proceeded to select the '2nd' number from each set, until I had five.

**Bhaal85**)I labelled them from 1-10 too, but then realised that I should have done 0-9, so I changed it, lol. It threw me quite a bit that question, how did people answer it? As this is something we don't normally do. I mean, what the hell, tables of random numbers? People pay for this crap? How does it work? lol. I then proceeded to select the '2nd' number from each set, until I had five.

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#53

I don't even know what a random no table is...I've never even heard of one before. I was so mighty confused when I got to that question. I just labeled each person with a number...in the order the numbers were given, then picked the first 5. tis about as wrong as you can get...but i figured doing something was better than nothign at all

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#54

Yeah, is that random number table stuff even on the syllabus, I've never seen it before, i just assigned each person with a range like 0 - 99 etc and then picked the first 5 numbers

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#55

yeah it is in the syllabus, its in the book somewhere in chapter 4 i think. its the bit with lots and lots of writing and u think, ill look at it later, and then dont. usually the question about number tables just come up in past papers about how u wud select a random sample, and u just vaguely mention it...not so this time

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#56

Well it was only 3 marks, and theres bound to be method marks in there somewhere, so whatever happened it cant be that bad

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#57

(Original post by

how did everyone do q6b, did u get lamde as 1.05 and then use the poisson formula to find x=0, x=1 x=2 and then do 1 minus that answer

**starcrossed**)how did everyone do q6b, did u get lamde as 1.05 and then use the poisson formula to find x=0, x=1 x=2 and then do 1 minus that answer

I personally approximated them to a binomial distribution, with N 50 and P 0.021. Then did 1 - p(x = 1,2,3).

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#58

(Original post by

Well it was only 3 marks, and theres bound to be method marks in there somewhere, so whatever happened it cant be that bad

**pi is exactly 3**)Well it was only 3 marks, and theres bound to be method marks in there somewhere, so whatever happened it cant be that bad

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#59

**Bhaal85**)

I labelled them from 1-10 too, but then realised that I should have done 0-9, so I changed it, lol. It threw me quite a bit that question, how did people answer it? As this is something we don't normally do. I mean, what the hell, tables of random numbers? People pay for this crap? How does it work? lol. I then proceeded to select the '2nd' number from each set, until I had five.

Divide the numbers you are given by 100. Round them UP to the nearest whole integer. Ignore repeats.

Therefore 079 => 0.79 approx = 1 (Choose surname A)

831 => 8.31 approx = 9 (Choose surname I)

etc...

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#60

(Original post by

I personally approximated them to a binomial distribution, with N 50 and P 0.021. Then did 1 - p(x = 1,2,3).

**Urdegish**)I personally approximated them to a binomial distribution, with N 50 and P 0.021. Then did 1 - p(x = 1,2,3).

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