OCR S2 (non-mei)

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Reply 100

lllllllllll

Does anyone remember Q7 (ii) by any chance? ie the actual question. I don't really remember any of it :confused:

(edited 8 years ago)

Reply 101

AMDD

Original post by Archer61

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?

Oh of course! Thank you i get it! How many marks do you reckon i lost for this? 2?

Reply 102

Younjaxx

Original post by AMDD

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

because (-a)^4 is just a^4 so it would be (1/2a^3)*((1/4)a^4 - +(1/4)a^4)
In the variance calculations you have (-a)^5 which gives you -a^5 because of the odd power but in the E(X) you have an even.

Reply 103

rich1334

Original post by Archer61

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?

4+5+1 marks I think. Either 10 or 11

Reply 104

ResultInExcellence

does anyone remember the order of accept/reject answers? wasn't able to store those numerical values onto my calculator, but I think I ended up getting something like accept, reject, accept?

Reply 105

tokopoko

Original post by ResultInExcellence

does anyone remember the order of accept/reject answers? wasn't able to store those numerical values onto my calculator, but I think I ended up getting something like accept, reject, accept?

the one involving the doctor was reject. the doctor was correct to say the mean has changed.

the others were accepts i think, hope this helps

Reply 106

lllllllllll

Original post by ResultInExcellence

does anyone remember the order of accept/reject answers? wasn't able to store those numerical values onto my calculator, but I think I ended up getting something like accept, reject, accept?

I assume you're talking about Q8 (ii)? If so I think all you need to do was model it using a binomial distribution

EDIT: Nvm, misinterpreted your post :tongue:

Reply 107

totoro1997

Does anyone know how to answer question 4???!!!! I thought it was impossible :frown:

Reply 108

ResultInExcellence

Original post by tokopoko

the one involving the doctor was reject. the doctor was correct to say the mean has changed.

the others were accepts i think, hope this helps

Yeah, I remember that one being reject as well. Well this seems to match so hopefully those are all correct!

Reply 109

ResultInExcellence

Original post by lllllllllll

I assume you're talking about Q8 (ii)? If so I think all you need to do was model it using a binomial distribution

EDIT: Nvm, misinterpreted your post :tongue:

Aha sorry, to be fair, I did phrase it very badly, but couldn't think of a better way of doing so xD

Reply 110

bentuk

pretty sure the probability is 0.6533 for 8ii), quite a few of my class got that, using different methods as well! Did anyone else get that?

Reply 111

AMDD

What will be the grade boundary for an A this year do you think?

Reply 112

Rifleboy123

Original post by lllllllllll

I think the continuity correction had to be -0.5 since we were finding P(X<x)

I'm like 80% sure it was less than or equal to, not just less than.

Reply 113

ResultInExcellence

Does anyone remember what they wrote for that 1 marker along the lines of whether it was necessary to have assumed that it was normally distributed?

Reply 114

irbif

Original post by AMDD

What will be the grade boundary for an A this year do you think?

i rekon itll be a=61 b=54 c=48 d=42 e=36 at a guess

Reply 115

tokopoko

* 8 ii) in full from memory* (checking my answer so might as well post it here)

Find probability that out of 4 tests at least 1 results in a Type 2 error. Probability of p=0.3 or 0.5 or 0.7 is 1/3.

P(type 2 when p=0.3) = 0
P(type 2 when p=0.5) = 0.6047 (using tables) NB (from part i) we know critical region X >= 8
P(type 2 when p=0.7) = 0.0933 (using tables)

P(Type 2 error) = 1/3 x 0.6047 + 1/3 x 0.0933 = 0.23267

B(4 , 0.23267)
P(X at least 1) = 1 - P(X=0) = 1 - (1 - 0.23267)^4

Ans = 0.6533

Reply 116