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# OCR MEI S1 Exam watch

1. Hey, I thought it was quite an easy paper compared to m1 exam on wedn. ))

I didn't have time to write down all the answers... the ones i have are:

q1, sd=23.31
iqr=25.5 (upper was around 20 and lower was negative)

q2, part iv) 0.064

q4, 2 tail test, 2.5% on each tail. for critical regions i got x>=7

How did it go for you guys?
2. Hey,

I got exactly the same critical region as u as well as IQR and s.d.. I found that exam quite difficult overall.

What did you put for q4 last part? I thought left tail didn't exist....well, i calculated that 0/12 you still accept H0. Therefore, i put that it's impossible to have left tail because you can't have negative no of packets or something like that.

i reckon i might have scraped an A, i bullsed up end of question 2 (such a retard not to set mean=variance even though i did initially then crossed it out). Everywhere else, although i was guessing at stuff, i reckon i was probably right. So A or B, but I really would like an A, don't want to blemish my record.

I shouldnt have spent so much time on m2 - spent hardly any time revising for s2, BAH!
5. (Original post by Willa)

i reckon i might have scraped an A, i bullsed up end of question 2 (such a retard not to set mean=variance even though i did initially then crossed it out). Everywhere else, although i was guessing at stuff, i reckon i was probably right. So A or B, but I really would like an A, don't want to blemish my record.

I shouldnt have spent so much time on m2 - spent hardly any time revising for s2, BAH!
AHHHHHHHHHHHHHHHHHHHH s2...Crappest exam in the world possibly?!
6. There was no left tail as the number was greater than 0.025. Did u reject H0 or accept H0?
7. (Original post by Mario_j)
There was no left tail as the number was greater than 0.025. Did u reject H0 or accept H0?
Oi Oi Mario! The lovely katie eh? (sorry, msn convo)! - Anyway, S2 is indeed the crappest exam of the year!
8. i accepted Ho. I'm pretty sure you have to accept Ho when the probability of getting the value is bigger than sign. level.

I think it was 0.0687>0.025
9. good - if we've all done crap at s2, then hopefully you'll get marks added to normalise the results - yippy
10. (Original post by Willa)
good - if we've all done crap at s2, then hopefully you'll get marks added to normalise the results - yippy
I bloody well hope so! Still got P3 to do yet, and was hoping to do better in this so i wouldnt have to do so well in P3! Damn it!!!
11. I think i cocked up S1 . Oh well.
12. (Original post by maljosh)
i accepted Ho. I'm pretty sure you have to accept Ho when the probability of getting the value is bigger than sign. level.

I think it was 0.0687>0.025
Didn't go too badly, much much much better than M1. In the last question I accepted Ho too. My answer to the last part was 28 packets, had to use logs. She needed to test 28 packets to get a critical region at the lower end. That is (I'm waffling i know) if she tested 28 packets and got no prizes you'd reject Ho and accept H1.
13. hi Jeremy, can you remember any of the answers from q2?
14. (Original post by JeremyG)
Didn't go too badly, much much much better than M1. In the last question I accepted Ho too. My answer to the last part was 28 packets, had to use logs. She needed to test 28 packets to get a critical region at the lower end. That is (I'm waffling i know) if she tested 28 packets and got no prizes you'd reject Ho and accept H1.
I got 17
The probability was 4/5 to get nothing, so
(4/5)^x = 0.025
x = ln(0.025)/ln(4/5) = 16.53

Or did i do it wrong?
15. (Original post by maljosh)
hi Jeremy, can you remember any of the answers from q2?
I got
i) 0.36
ii) 0.504
iii) 0.1152
iv) 0.317
v) 0.2
16. (Original post by maljosh)
hi Jeremy, can you remember any of the answers from q2?
Question 2.
P for B =0.4
P for A = 0.6

0.064, same as you

Find the probablilty of A winning one frame only and B winning
then if A wins one frame
ABBB
BABB
BBAB
Probability for each combination = 0.0384
multiplied by 3 gives : 0.1152

Show the probability of B winning is 0.317

AABBB
ABABB
ABBAB
BAABB
BABAB
BBAAB
Probability for each combination = 0.02304
muliplied by 6 gives = 0.13804

Add all three together, 0.064+0.1152+0.13804 = 0.317

Just trying to remember the rest.............
17. Given that B wins, find the probability he wins in three frames:

0.064/0.317 = 0.202
18. (Original post by JamesF)
I got 17
The probability was 4/5 to get nothing, so
(4/5)^x = 0.025
x = ln(0.025)/ln(4/5) = 16.53

Or did i do it wrong?
Yep,

0.8^x = 0.025
xlog0.8 = log0.025
x=log0.025/log0.8
x= 16.53

Not quite sure where I got 28 from......................
19. for part iv) i think you have to consider all possible winning situations like,

BBB = 0.4^3=0.064

for BBB only 1 possible way 3!/3!=1

ABBB=(0.6*0.4^3)*3=0.1152

for ABBB etc there are 4!/3! -1 = 3 ways

AABBB etc=(0.6^2*0.4^3)*8=0.02304*6=0. 13824

for AABBB= 5!/2!3! -4= 6 ways

P(win)=0.064+0.1152+0.13824=0.31 744 or 0.317

I think that's how i've done it....
20. for conditional probability i'm not quite sure what i've done...

i'm sure i've used the correct formula... but can't remember getting 0.202,

i think i got a slightly lower result... anyways, i'll get method marks,

So for that paper i'll get at least a B.

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