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Edexcel S2 - 27th June 2016 AM Watch

  • View Poll Results: How did you find the Edexcel S2 exam?
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    (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.
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    (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)
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    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
  1. File Type: pdf Question.pdf (10.6 KB, 98 views)
  2. File Type: pdf Mark Scheme.pdf (106.2 KB, 96 views)
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    (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
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    (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
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    (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.
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    (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
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    (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.
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    How many marks was the P(7<R<10) question
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    (Original post by igotohaggerston)
    How many marks was the P(7<R<10) question
    Two
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    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.
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    (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
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    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

    AND ANSWER:
    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
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    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 ?
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    (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
    not sure about this one someone please reply to this one: 23.69%>1% accept H0.

    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
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    (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
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    (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.
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    does anyone think i am wrong on hypothesis testing q i got 23.69% >1% so accept H0?
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    (Original post by igotohaggerston)
    wow are you sure I thought it was 2 or 3 marks
    No it was definitely 1 mark
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    (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

    AND ANSWER:
    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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