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# Edexcel S2 - 27th June 2016 AM

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• View Poll Results: How did you find the Edexcel S2 exam?
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1. (Original post by asacred)
can you please explain how did you get that ? instead of just saying "ur wrong "
You had to work out E(A)=E(piR^2)=piE(R^2) the reason you got 49pi is because you thought that E(R^2)= (E(R))^2 which it doesnt it is (E(R))^2+var(R) or just integrate it funny because this question is in the s2 book as an example but with different numbers obviously.
2. (Original post by igotohaggerston)
You had to work out E(A)=E(piR^2)=piE(R^2) the reason you got 49pi is because you thought that E(R^2)= (E(R))^2 which it doesnt it is (E(R))^2+var(R) or just integrate it funny because this question is in the s2 book as an example but with different numbers obviously.
ikr, i got it right, you quoted the wrong person x)
3. For all of those still debating wherever the answer is 151/3 pi or 49 pi,

Here I've attached a similar past paper question and the mark scheme as the question we've done today.

I only done this yesterday so it prepared me very well for today :P.
Attached Images
4. Question.pdf (10.6 KB, 63 views)
5. Mark Scheme.pdf (106.2 KB, 69 views)
6. (Original post by asacred)
squarring e(R) does not equal E(R²)....
Honestly I dont remember what I did but I might havae got it wrong looking back on it and you might be right. I was going to do it the other way (VAR) way but I didnt have enough time so I had to choose between the one way :P
7. (Original post by igotohaggerston)
You had to work out E(A)=E(piR^2)=piE(R^2) the reason you got 49pi is because you thought that E(R^2)= (E(R))^2 which it doesnt it is (E(R))^2+var(R) or just integrate it funny because this question is in the s2 book as an example but with different numbers obviously.
Lol yeah! I even looked at it in the morning such a shame still got it wrong. What marks do you think I would get for that Question
8. (Original post by fpmaniac)
Pretty sure it was 49pi and i got other answer for n=225, I got n=200 something
R-U[5,9]

A=piR"

E(A)=pi*E(R")

E(R)=(5+9)/2=14/2=7

VAR(R)=(9-5)"/12=16/12=4/3

VAR(R)=E(R")-[E(R)]"
E(R")=VAR(R)+[E(R)]"

E(R")=(4/3)+(7)"
E(R")=(4/3)+49

E(R")=151/3

E(A)=151pi/3
" means squared.
9. (Original post by ISAAQAHELP)
R-U[5,9]

A=piR"

E(A)=pi*E(R"

E(R)=(5+9)/2=14/2=7

VAR(R)=(9-5)"/12=16/12=4/3

VAR(R)=E(R"-[E(R)]"
E(R"=VAR(R)+[E(R)]"

E(R"=(4/3)+(7)"
E(R"=(4/3)+49

E(R"=151/3

E(A)=151pi/3
" means squared.
Yeah I realised where I went wrong. Oh well
10. (Original post by ISAAQAHELP)
R-U[5,9]

A=piR"

E(A)=pi*E(R"

E(R)=(5+9)/2=14/2=7

VAR(R)=(9-5)"/12=16/12=4/3

VAR(R)=E(R"-[E(R)]"
E(R"=VAR(R)+[E(R)]"

E(R"=(4/3)+(7)"
E(R"=(4/3)+49

E(R"=151/3

E(A)=151pi/3
" means squared.
Yup. Perfect.
11. How many marks was the P(7<R<10) question
12. (Original post by igotohaggerston)
How many marks was the P(7<R<10) question
Two
13. Would I get any marks for this?! Question 5) Normal approx...X~B(n,0.2)E(X)=0.2nVar( X)=0.16ntherefore; X~N(0.2n,0.16n) P(X>55)=> CC P(X>=55.5)P(Z>(55.5-0.2n)/rt0.16n)=0.04011-P(Z<(55.5-0.2n)/rt0.16n)=0.0401P(Z<(55.5-0.2n)/rt0.16n)=0.9599Therefore;(55.5-0.2n)/rt0.16n = 1.7555-0.2n= 1.75(0.4)(n^1/2)55-0.2n=0.7n^1/2Then I squares out formed a quadratic and tried to rearrange but it didn't solve (yielded imaginary numbers) How many marks out of 8 would I get if any? Was I on the right lines or have I lost the plot.
14. (Original post by SB0073)
Would I get any marks for this?! Question 5) Normal approx...X~B(n,0.2)E(X)=0.2nVar( X)=0.16ntherefore; X~N(0.2n,0.16n) P(X>55)=> CC P(X>=55.5)P(Z>(55.5-0.2n)/rt0.16n)=0.04011-P(Z<(55.5-0.2n)/rt0.16n)=0.0401P(Z<(55.5-0.2n)/rt0.16n)=0.9599Therefore;(55.5-0.2n)/rt0.16n = 1.7555-0.2n= 1.75(0.4)(n^1/2)55-0.2n=0.7n^1/2Then I squares out formed a quadratic and tried to rearrange but it didn't solve (yielded imaginary numbers) How many marks out of 8 would I get if any? Was I on the right lines or have I lost the plot.
I guess 3
15. THIS IS THE QUESTION FOR THE normal APPROX.

X-B(N,0.2) USING NORMAL APPROX AND GIVEN THAT P(X>55)=0.0401
FIND N

X-B[N,0.2]= W-N[X,Y"] X IS MEAN Y" IS VARIANCE " MEANS SQUARED

P(X>55)=P[W>55.5)=P(Z>([55.5-X]/Y)=0.0401

P(Z<([55.5-X]/Y)=1-0.0401=0.9599

USING TABLE

(55.5-X)/Y=1.75
55.5-X=1.75Y
1.75Y+X=55.5 NOW USING THE BINOMIAL PARAMETERS
WE CAN FIND Y AND X IN TERMS OF N
X=0.2N
Y"=0.2*0.8N Y"=0.16N Y=0.4N! WHERE !=SQUARE ROOT
SO 1.75(0.4N!)+0.2N =55.5
0.7N!+0.2N=55.5
0.2N+0.7N!-55.5=0 NOW SOLVE THIS USING QUADRATIC FORMULA
YOU GET N!=15 OR (-18.5)
SO N = 225 OR 342.25 BUT WE ARE DEALING STUDENTS
SO N=225 ONLY
don't KNOW WHY PEOPLE ARE GETTING 200
16. Did anyone end up using logs for one of the questions - i can't remember the question but you had to find like the lowest value of Y or something and it involved the binomial ?
17. (Original post by ISAAQAHELP)
Q1-
1.41=1.4419 so mean=varaiance so use poisson
0.2510
0.7769
0.1378
0.1516

Q2-
n=60
0.0133
0.0159

Q3-
1/2
151PI/3 REPLY TO THIS as well PLZ

Q4-
B=8
A=-12
K=1/9

Q5-
N=225

Q6-
57

Q7-
1.2
0.24
41.875 AND 40 REPLY TO THIS ONE IT IS ABOUT THE MONEY LAST ONE.
Could you remind me of what the question was to the answer 0.7769
18. (Original post by DesignPredator)
Could you remind me of what the question was to the answer 0.7769
can't remember lol just typed in my calculator so i can keep it there till i get home to check answers
19. (Original post by ISAAQAHELP)
can't remember lol just typed in my calculator so i can keep it there till i get home to check answers
I got it looking at the tables it must be what is the probability of at least 1 cherry.
20. does anyone think i am wrong on hypothesis testing q i got 23.69% >1% so accept H0?
21. (Original post by igotohaggerston)
wow are you sure I thought it was 2 or 3 marks
No it was definitely 1 mark
22. (Original post by ISAAQAHELP)
THIS IS THE QUESTION FOR THE normal APPROX.

X-B(N,0.2) USING NORMAL APPROX AND GIVEN THAT P(X>55)=0.0401
FIND N

X-B[N,0.2]= W-N[X,Y"] X IS MEAN Y" IS VARIANCE " MEANS SQUARED

P(X>55)=P[W>55.5)=P(Z>([55.5-X]/Y)=0.0401

P(Z<([55.5-X]/Y)=1-0.0401=0.9599

USING TABLE

(55.5-X)/Y=1.75
55.5-X=1.75Y
1.75Y+X=55.5 NOW USING THE BINOMIAL PARAMETERS
WE CAN FIND Y AND X IN TERMS OF N
X=0.2N
Y"=0.2*0.8N Y"=0.16N Y=0.4N! WHERE !=SQUARE ROOT
SO 1.75(0.4N!)+0.2N =55.5
0.7N!+0.2N=55.5
0.2N+0.7N!-55.5=0 NOW SOLVE THIS USING QUADRATIC FORMULA
YOU GET N!=15 OR (-18.5)
SO N = 225 OR 342.25 BUT WE ARE DEALING STUDENTS
SO N=225 ONLY
don't KNOW WHY PEOPLE ARE GETTING 200
I've done something very similar to you but I appear to have made a silly mistake, resulting in an answer of 124 or something from my quadratic formula. What do you think I might have lost in terms of marks?

(Original post by fefssdf)
Did anyone end up using logs for one of the questions - i can't remember the question but you had to find like the lowest value of Y or something and it involved the binomial ?
Yes, I took ln of both sides and it appears I got the right answer, hope this is an OK method to use.. should be given that S2 is usually taught alongside A2 maths which includes ln in C3 and C4.

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