You are Here: Home >< Maths

# OCR S2 (non-mei) watch

1. (Original post by Molly_xox)
Yeah, +0.5...

38-50/sq rt(50) = -1.697, and I thought you needed to compare it to -1.645?
Isnt the test supposed to be (X+0.5)-50/rt(50) ? So you'd 38.5-50/rt(50) ?

I'm not 100% sure, you may very well be right
2. (Original post by Molly_xox)
Yeah, +0.5...

38-50/sq rt(50) = -1.697, and I thought you needed to compare it to -1.645?
I think the continuity correction had to be -0.5 since we were finding P(X<x)
3. can someone please write out the 2 simultaneous equations they got for question 1, i think i made a very stupid error
thanks!
4. (Original post by tokopoko)
I felt pretty good about all my answers, ill post as many as i can remember.

a=6 then a=1.5 for cont dist qn

Mean = 146, S.D=2 for normal qn

Very last qn was something like:
1/3(0.6 + 0.03) <-- i really cant remember
So P(type 2 error) = 0.23 ish

So P(X at least 1 out of 4) = 1 - P(X=0) = 1- (1-0.23)^4
Yep I got exactly the same as you for all the questions.
7ii I got 37 (critical value was like 38.4 and then continuity correction so its 37.9 and then you round down to 37)
5. (Original post by lllllllllll)
I think the continuity correction had to be -0.5 since we were finding P(X<x)
I thought we were finding P(X less than or equal to x) ?
6. (Original post by lllllllllll)
I think the continuity correction had to be -0.5 since we were finding P(X<x)
ye i thought so too. i.e. P(X<38) would change to P(X<37.5)?
7. Can someone give the equation for question 6) I think I looked at the equation wrong because I got a = 1/6 and 6
8. (Original post by Archer61)
Can someone give the equation for question 6) I think I looked at the equation wrong because I got a = 1/6 and 6
Think it was something like 1.5a^-3 * x^2
9. (Original post by Younjaxx)
Yep I got exactly the same as you for all the questions.
7ii I got 37 (critical value was like 38.4 and then continuity correction so its 37.9 and then you round down to 37)
please explain why you got 37.9, i thought C.Correction was done before not after the test
10. just realised for the first one I did it as sigma^2, when the formula means it'd be sigma, so I got sigma^2 = 2 and thus sigma = sqrt(2)
would I lose a mark for this? Suppose I would do, looking at a past paper with a similar question doesn't seem to allow sigma^2
11. (Original post by AMDD)
What did everyone get as the value for a for each part of the questions? I got a=6 for both....
part two was 1.5 and I know the mistake you made because I made the same mistake, when calculating E(X) it was 1/(2a^3) * ( (1/4)x^4) between a and -a so if you put that in you get 1/(2a^3) 8 ((1/4)a^4 - 1/4a^4) the mistake you made was to put ((1/4)a^4 + ((1/4)a^4) but remember that -a^4 is a^4 so if you integrate it you get E(X) as 0. hope it makes sense
12. (Original post by rich1334)
I thought we were finding P(X less than or equal to x) ?
(Original post by tokopoko)
ye i thought so too. i.e. P(X<38) would change to P(X<29.5)?
Possible...I don't remember Q7 off the top of my head. All I remember is a continuity correction with -0.5
13. (Original post by tokopoko)
please explain why you got 37.9, i thought C.Correction was done before not after the test
Because you're working backwards. So what I did was say ok so what value of x gives us z of 0.05 and that gave me x as 38.4 or something. And then I was like ok if we're saying that x=38.4 is the critical region, we must've had 37.9 which we then corrected to 38.4 so the actual value of x is 37.9 and the answer is 37. So imagine you wanted to work out the P(x=<38) you would actually work out P(x=<38.5) so if you have P(x=<38.4) the original thing you wanted to find out was P(x=<37.9) makes sense?
14. were the 2 simultaneous equations for question 1:
150-u=2(sigma)
and
u-146=2(sigma)
thanks
15. (Original post by rich1334)
Isnt the test supposed to be (X+0.5)-50/rt(50) ? So you'd 38.5-50/rt(50) ?

I'm not 100% sure, you may very well be right
I hadn't even thought about that...but then you can't have half a signal failure?

(Original post by lllllllllll)
I think the continuity correction had to be -0.5 since we were finding P(X<x)
Yeah that might be where I went wrong, I thought it was strange that I was rounding up...
16. (Original post by Younjaxx)
part two was 1.5 and I know the mistake you made because I made the same mistake, when calculating E(X) it was 1/(2a^3) * ( (1/4)x^4) between a and -a so if you put that in you get 1/(2a^3) 8 ((1/4)a^4 - 1/4a^4) the mistake you made was to put ((1/4)a^4 + ((1/4)a^4) but remember that -a^4 is a^4 so if you integrate it you get E(X) as 0. hope it makes sense
Thatnkyou so much for replying and i think i followed that but why would it not be (1/2a^3)*((1/4)a^4- - (1/4)a^4) inside the brackets which is where i seem to have gone wrong
17. (Original post by Younjaxx)
Because you're working backwards. So what I did was say ok so what value of x gives us z of 0.05 and that gave me x as 38.4 or something. And then I was like ok if we're saying that x=38.4 is the critical region, we must've had 37.9 which we then corrected to 38.4 so the actual value of x is 37.9 and the answer is 37. So imagine you wanted to work out the P(x=<38) you would actually work out P(x=<38.5) so if you have P(x=<38.4) the original thing you wanted to find out was P(x=<37.9) makes sense?
hmm, ive never seen that before. you may be right but i would never think to do that. hopefully someone else can confirm
18. (Original post by thickmaster)
I stored all my answers in my calculator and here they are (I think 8ii is wrong):
1. Mean = 146, sigma = 2
2. i) The rabbits occur independent of each other
ii) 0.222 (Using poisson)
3. i) a = 6
ii) a = 1.5
4. 0.8311
5. 0.2536
6. 0.9807
7. i) 0.1247
ii) 38
8. i) 0.0316
ii) 0.961 (I did 1 - (1/3((1-0.9685)+(1-0.6047)+(1-0.0933)))^4)
I think 8ii is wrong as the probability of getting a type II error when p=0.3 was 0 as you would've accepted H0 correctly rather than erroneously so there was no error.

Posted from TSR Mobile
19. (Original post by AMDD)
Thatnkyou so much for replying and i think i followed that but why would it not be (1/2a^3)*((1/4)a^4- - (1/4)a^4) inside the brackets which is where i seem to have gone wrong
Substituting -a into x^4 gives you -a*-a*-a*-a giving you positive a^4.

Also anyone remember how many marks question 3 was worth?
20. (Original post by iwishiwasclever)
were the 2 simultaneous equations for question 1:
150-u=2(sigma)
and
u-146=2(sigma)
thanks
150 - U = 2 sigma
146 - U = -1.5 sigma pretty sure they didnt both have 2 in

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: June 13, 2015
Today on TSR

### AQA Mechanics Unofficial Markscheme

Find out how you've done here

### 2,809

students online now

Exam discussions

### Find your exam discussion here

Poll
Useful resources

Can you help? Study help unanswered threadsStudy Help rules and posting guidelinesLaTex guide for writing equations on TSR

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE