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    (Original post by MahuduElec)
    Actually! sorry to say but it was 16.7 100%. VarX = 12.26 -3.08^2 = 2.7736

    VarF=Var(10X)=100Varx(X)= 100 x 2.7736

    SD = root(277.36) = 16.7 3sf, this is exactly what I did know that i remember the 16.65
    O of course!

    Damn that rule of Var(aX) = a^2Var(X)!

    (Original post by sam_97)
    It 100% asked for standard deviation I think it was 18.2 not 18.7, but yeah 18.2 lied inside the confidence interval
    Yes you're right!

    Phew that was lucky!
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    (Original post by MahuduElec)
    Actually! sorry to say but it was 16.7 100%. VarX = 12.26 -3.08^2 = 2.7736

    VarF=Var(10X)=100Varx(X)= 100 x 2.7736

    SD = root(277.36) = 16.7 3sf, this is exactly what I did know that i remember the 16.65

    I think your mistake was doing VarF=10VarX, it was 10^2VarX
    I got the same as you
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    (Original post by Suits101)
    Oh I see what you mean...

    It was a straight line but you wouldn't treat it as a rectangular distribution.

    In my opinion marks are for:

    Method 1: splitting into shapes with integration

    Area of triangle (1)
    Integrating second function with limits (1) and correct integral with limits substituted (1)
    Final answer (1)

    Method 2: pure integration

    Integrating function 1 with limits (1)
    Integrating function 2 with limits (1)
    Correct integration and substitution of limits (1)
    Final answer (1)
    Hm okay then buddy. I'm hopeful to get at least 1 if not 2 of those method marks then as I did integrate. Cheers! :-)
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    (Original post by Suits101)
    Do you know how many marks these were worth?

    1(e) - last part of Q1
    3(b) - forgot to multiply by 2, book question
    3(e)(ii) - I got 5.27p but apparently it asked for variance not SD

    Also (sorry!) what did you put about Gerald's claim? I said CI is an element of 18.7 therefore no valid/proper conclusion can be made - also what did you do for last part of 6 as I just calculated a mean and said it doesn't support belief because answer was > 18.7?
    1E was 3 marks, you'll lose 1 on 3b (I did this too) and probably 1 on 3eii

    I said his claim cannot be supported because of exactly what you said
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    I think I got about 60, any guesses on what UMS this will be??


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    (Original post by luke010203)
    I think I got about 60, any guesses on what UMS this will be??


    Posted from TSR Mobile
    mid to high 70s I'd guess
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    unofficial markscheme

    1a. 0.703 (1)
    b.0.769 (3)
    c.0.133 (3)
    d.0.144 (3)
    e.0.268 (3)

    2. mean =31.7 sd=1.719 (4)

    3a. 0.35 (1)
    b. 0.158 (3)
    c. 3.08 (4)
    d. 2.77 (2)
    e. mean dosent = variance and something else (1)
    f. mean= 31p sd=17p (3)

    4a. k=10 (1)
    b. 0.7 (1)
    c.0.05 (2)
    d.0.0289 (3)

    5a. chi value =2.3388 reject Ho (11)
    b. some wordy stuff (2)

    6a, -2.08 accept HA (7)
    b. (14.9, 18.55) (7)
    c. wordy stuff (2)

    7. differentiate and plot (5)
    b. mean=3 (4)
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    for the 2 people choosing books, did a lot of people not times the answer by 2? for 3 marks I'd be surprised if you didn't have to times it by anything. My logic was that there are 2 people, so 2 combinations, I used a tree diagram to confirm.
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    (Original post by AngelicAmazing)
    1E was 3 marks, you'll lose 1 on 3b (I did this too) and probably 1 on 3eii

    I said his claim cannot be supported because of exactly what you said
    I think I may lose 2 on 3(e)(ii)?

    I'm thinking marks are for:

    0.45 identified
    0.35 x 0.45 x 2
    Final answer
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    (Original post by Busted838)
    unofficial markscheme

    1a. 0.703 (1)
    b.0.769 (3)
    c.0.133 (3)
    d.0.144 (3)
    e.0.268 (3)

    2. mean =31.7 sd=1.719 (4)

    3a. 0.35 (1)
    b. 0.158 (3)
    c. 3.08 (4)
    d. 2.77 (2)
    e. mean dosent = variance and something else (1)
    f. mean= 31p sd=17p (3)

    4a. k=10 (1)
    b. 0.7 (1)
    c.0.05 (2)
    d.0.0289 (3)

    5a. chi value =2.3388 reject Ho (11)
    b. some wordy stuff (2)

    6a, -2.08 accept HA (7)
    b. (14.9, 18.55) (7)
    c. wordy stuff (2)

    7. differentiate and plot (5)
    b. mean=3 (4)
    Lot's of stuff wrong on this..
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    (Original post by Busted838)
    unofficial markscheme

    1a. 0.703 (1)
    b.0.769 (3)
    c.0.133 (3)
    d.0.144 (3)
    e.0.268 (3)

    2. mean =31.7 sd=1.719 (4)

    3a. 0.35 (1)
    b. 0.158 (3)
    c. 3.08 (4)
    d. 2.77 (2)
    e. mean dosent = variance and something else (1)
    f. mean= 31p sd=17p (3)

    4a. k=10 (1)
    b. 0.7 (1)
    c.0.05 (2)
    d.0.0289 (3)

    5a. chi value =2.3388 reject Ho (11)
    b. some wordy stuff (2)

    6a, -2.08 accept HA (7)
    b. (14.9, 18.55) (7)
    c. wordy stuff (2)

    7. differentiate and plot (5)
    b. mean=3 (4)
    Hi,

    3(e) I don't think they'll accept mean =/= variance as this was given in stem of question (they were similar, similar means it can be used) hence I think asnwers should be that people don't borrow books at a constant average rate and Poisson distribution has no upper limit but model X does (6 books)
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    (Original post by MahuduElec)
    for the 2 people choosing books, did a lot of people not times the answer by 2? for 3 marks I'd be surprised if you didn't have to times it by anything. My logic was that there are 2 people, so 2 combinations, I used a tree diagram to confirm.
    I didn't times by 2. Urgh. I won't get the answer mark or the method mark for x2. Does that mean I'll get 1 for working out 0.35 and 0.45 and x them together?
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    For Poisson, would e^-lambda=/0 for P(X=x) be allowed as the second B1? The other being the upper limit.
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    (Original post by MahuduElec)
    for the 2 people choosing books, did a lot of people not times the answer by 2? for 3 marks I'd be surprised if you didn't have to times it by anything. My logic was that there are 2 people, so 2 combinations, I used a tree diagram to confirm.
    Yeah man, you have to times by 2 as there are 2! combinations as you said.
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    Hi I'll try an unofficial markscheme:

    1
    (a) 0.70291
    (b) 0.6691
    (c) 0.13271
    (d) 0.1441
    (e) 0.622

    2. Mean = 31.7 s2 = 2.96

    3
    (a) P(X > 3) = 0.3503
    (b) P (X < 3) and P (X > 3) = 0.3153
    (c) Show E(X) = 3.08, Var(x) = 2.77363
    (d) Poisson distribution cannot have an upper limit, people don't borrow books at a constant average rate
    (e)(i) Mean = 30.8p
    (e)(ii) Standard deviation = 16.7p

    4
    (a) k = 10
    (b) P = 0.70
    (c) E(X) = 0.05
    (d) Integrate, ans = 1/300
    (e) SD(X) = 0.02895

    5
    (a) Test statistic = 2.3386, CV = 2.706 so accept null
    (b) comment

    6
    (a) Test statistic = -2.20794, reject null
    (b)(i) 16.7 +- 1.80
    (b)(ii) Claim not valid because 18.2 lies in CI
    (c) -

    7
    (a) Draw graph - DENSITY function, so have to differentiate
    (b) 73/24
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    (Original post by KB_97)
    I didn't times by 2. Urgh. I won't get the answer mark or the method mark for x2. Does that mean I'll get 1 for working out 0.35 and 0.45 and x them together?
    If the mark scheme is consistent with past papers, then yes you will
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    (Original post by Busted838)
    unofficial markscheme

    1a. 0.703 (1)
    b.0.769 (3)
    c.0.133 (3)
    d.0.144 (3)
    e.0.268 (3)

    2. mean =31.7 sd=1.719 (4)

    3a. 0.35 (1)
    b. 0.158 (3)
    c. 3.08 (4)
    d. 2.77 (2)
    e. mean dosent = variance and something else (1)
    f. mean= 31p sd=17p (3)

    4a. k=10 (1)
    b. 0.7 (1)
    c.0.05 (2)
    d.0.0289 (3)

    5a. chi value =2.3388 reject Ho (11)
    b. some wordy stuff (2)

    6a, -2.08 accept HA (7)
    b. (14.9, 18.55) (7)
    c. wordy stuff (2)

    7. differentiate and plot (5)
    b. mean=3 (4)
    Your book answer is wrong it should be 0.45 * 0.35 * 2 as there were two arrangements.
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    (Original post by Suits101)
    Hi I'll try and unofficial markscheme:
    1(a) 0.70291
    (b) 0.6691
    (c) 0.13271
    (d) 0.1441
    (e) 0.622

    Mean = 31.7 s2 = ?

    3(a) 0.3503
    (b) 0.3153
    (c) Var(x) = 2.77363
    (d) Poisson cannot have an upper limit, people don't borrow books at a constant average rate
    3(e)(i) 30.8p
    3(e)(ii) 16.7p

    4(a) k = 10
    (b) P = 0.70
    (c) E(X) = 0.05
    (d) Integrate, ans = 1/300
    (e) SD(X) = 0.02895

    (a) Test statistic = 2.3386, CV = 2.706 so accept null5(b) comment6(a) Test statistic = -2.20794, reject null6(b)(i) 16.7 +- 1.806(b)(ii) Claim may be valid because 18.7 lies in CI6(c) -7(a) Draw graph - DENSITY function, so have to differentiate7(b) 73/24
    Only disagree with 6Bii, the Ci was 14.9-18.5, the mainland mean was 18.2, which lies within the CI so the Ci does not support Gerald's claim
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    (Original post by MahuduElec)
    Only disagree with 6Bii, the Ci was 14.9-18.5, the mainland mean was 18.2, which lies within the CI so the Ci does not support Gerald's claim
    Sorry that's what I meant check edit
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    (Original post by Suits101)
    Sorry that's what I meant check edit
    Nice format, However you sill left it as claim may be vaild
 
 
 
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