OCR MEI S1 Maths - 26th May 2011 - Question Paper

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all of these pages are the solutions of s1 june paper 2011. I have done the section a of s1. If you want section b, u will have to wait by tommorow. I don't have much time to do it becuase i have to prepare for my a2 ict tommorow. I have done s1 last year and i got 97/100 on the mei maths. At the moment i am doing s2 for my further maths, ive done my maths for a2 and as combines last year . You might find it very useful.

Reply 21

Funkyhouse

Original post by chris_gloyne

no they will have a range of what it can be so you will probably get the marks.

Thank you

Reply 22

css-

4 i) would I be penalised for putting the 0 line up the axis, or does there have to be a gap?

also the probability arguments for the dice, would it be okay to do it by words rather than the two-way tables providing i was clear in my explanation?

Reply 23

knabber99

Original post by Aaqil

Do you think I'd lose a mark for doing the y-axis as a common denominator of 18? e.g 1/9 as 2/18 and so on? :s-smilie:

Reply 24

TimetoSucceed

guys is 59/72 an A? =/

Reply 25

css-

Original post by knabber99

Do you think I'd lose a mark for doing the y-axis as a common denominator of 18? e.g 1/9 as 2/18 and so on? :s-smilie:

No, I did that and confident it will be fine as long as all the conversions to /18 are correct :smile:

Reply 26

css-

Original post by TimetoSucceed

guys is 59/72 an A? =/

Could be, usually around the 60 mark and a lot of people seem to have found this quite difficult.

Reply 27

Jordannn

For the midrange maximum and minimum you dont add or subtract 999 coz the data is continuous, you effectively add 999.99 which rounds to 1000. the midrange wont end it point 5 coz of this.

Reply 28

TimetoSucceed

for the part of P(A l B)

I did 0.11/0.3, then i gave the reasoning

its out of 3 marks, would I get any? =/

Reply 29

css-

Original post by Jordannn

For the midrange maximum and minimum you dont add or subtract 999 coz the data is continuous, you effectively add 999.99 which rounds to 1000. the midrange wont end it point 5 coz of this.

what were they then, 1500 and 2500?

Reply 30

TimetoSucceed

looks like i got 56-57/72

just my luck

Reply 31

Blutarsky

Original post by TimetoSucceed

looks like i got 56-57/72

just my luck

Same. That's what I want though, so I'm not too bothered.

Reply 32

Jordannn

Original post by css-

what were they then, 1500 and 2500?

yup thats what i put :smile:

Reply 33

ScheduleII

Original post by jimblue

question 8iv) A and B are correct

no 8iv)b) is incorrect.
P (X>9.5)= 0.76

P (At least 2 of 3) = 0.76 ^3 which is 8iv)a)
plus 3C2 (0.76^2) (1-0.76) for 2 sacks of coal have heating quality above 9.5
which comes up as 0.76^2 (0.72), not the same as 8iv)a) so b) cannot be double.
it's 0.416
so 8iv)b) is 0.855

Reply 34

css-

Original post by Jordannn

yup thats what i put :smile:

So did I, you're the first other person I've seen do it!

Reply 35

Jordannn

Original post by css-

So did I, you're the first other person I've seen do it!

Great minds think alike!!! but then again fools rarely differ... so we are either right or wrong! at least thats cleared up!

Reply 36

selby118

I got
1i. 13 ii. positive iii. 1000 and 2000
2i. 1/120 ii. 1/10
3i. 0.1780 ii. 8.8989
4ii A) double sided diagram showing 10/36 possible ways = 5/18
B) 6x(1/6 x 1/6) = 1/6
iii. 1.9444
5i. 0.03, 0.11, 0.3, 0.56
ii. P(WnF) = P(W) x P(F) if independent, so not independent as 0.11? 0.0574
iii. 0.11/0.41 = 0.2683, shows probability of working part time, given that they are a woman
6i. mean = 2.2143
s = 0.7782
ii. mean decreases
s increases
7i(A) 20C1 x 0.15 x 0.85 ^19 = 0.1368
(B) P(X is larger than or equal to 2) = 1 - P(X is smaller than or equal to 1) = 0.8244
ii. p = probability of no-show
H0: p=0.15
H1: p<0.15 (stated that it would reduce probability)
iii. X~B(20,0.15)
P(X smaller than or equal to 1) = 0.1756, not sufficient to reject H0
iv. 6<8, so is in critical region, so sufficient to reject H0
v. no values below 0.05 on tables, so never sufficient to reject H0
8ii. mean = 9.69, s = 0.3
iii. All data lies between LQ - (1.5 x IQR) and UQ + (1.5 x 0.3), 9.06 to 10.26
iv (A) (38/50) x (37/49) x (36/48) = 0.4078
(B) 3C2 x ((38/50) x (37/49) x (12/48)) + 0.4078 = 0.8382

Reply 37

Aaqil

Original post by knabber99

Do you think I'd lose a mark for doing the y-axis as a common denominator of 18? e.g 1/9 as 2/18 and so on? :s-smilie:

No you won't, both of these fractions are acceptable.

Reply 38

Aaqil

Original post by selby118

In question 6(ii) the answer that you put down must be wrong because the female gulls did not lay any eggs so the mean doesn't get affected and vice versa s.d. doesn't get affected too.

Reply 39

chingchong

yh but the female eggs would have frequency right? So therefor it would increase sigma f! So when u do the mean the sigma fx wouldnt be affected but the sigma f would be larger! Therefore sigma fx minus and increase sigma f would cause the mean to fall!