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    (Original post by kyle18)
    q1) a) 0.703 b) 0.967 (dunno if right) c) 0.133 d) 0.144 e) 0.622
    q2) I used Z so could be wrong as thinking needed to use t. mean= 31.7 var= 4.3264
    3a) 0.35 b) 0.1575 c) mean proof of 3.08, 2.7736 d) independent, possion has no top limit, random doesn't occur at a constant rate etc. e)i) 30.8 e)ii) 4.08 (i think this is what i got)
    4a) k=10 b) 0.7 c) 0.35 d) 1/300 e) (root3)/60
    5a) 2.3386 chi calc, 2.706 chi crit, hence accept H0. 5b) some logical explanation linking Observed and expected and then the suggested engine
    6a) accept H0 i think 6bi) 14.9 to 18.5 6bii) explanation 6c) mean is above 18.2 hence doesn't support his claim.
    7a) f(x) = 1/4 from x=1,2,3,4. then negative gradient from (4,1/4) to (6,0)
    7b) 73/24

    I believe most of these are right. 1b, 2, and 6bi i need conformation of these results then we'll have an unofficial mark scheme.
    grade boundaries i think will be
    A* 68-69 A 63-64 B 57-58 C 51-52
    Full UMS would be at 74/75 as easier than last year so would expect about 2 above. Hope this has helped.
    1(b) 0.669 worked out using 1 - [P(X=0) + P(X=1)] with lambda = 2.3

    6(b)(i) agree

    3(d) agree with all reasons but independent is not accepted as it's given in question (saying it will give you no mark but you won't lose a mark for saying it as it's still right)
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    (Original post by luke010203)
    For question 6 what were the values for the means of island A/B/C and their sample sizes? and what was the standard deviation you had to use in the confidence interval? I think it said it was unknown so had to use the t value


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    6(a) SD was given so z values for hypothesis test
    6(b)(i) SD was unknown so had to use t values for confidence interval
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    (Original post by sam_97)
    I think you might be thinking about the median there, the question was about the mean.

    (Original post by MahuduElec)
    That is the method for the median. since Fx is a cumulative function, Fx=0.5 gives the median value when solving for X
    :facepalm:
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    drew F(x) feelsbadman
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    (Original post by Suits101)
    6(a) SD was given so z values for hypothesis test
    6(b)(i) SD was unknown so had to use t values for confidence interval
    Can you remember the island data from the table??
    For island B I think it was 116.9 and n was 7, so the mean was 16.7 but I can't remember what the standard deviation was to calculate the confidence interval


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    (Original post by Suits101)
    1(b) 0.669 worked out using 1 - [P(X=0) + P(X=1)] with lambda = 2.3

    6(b)(i) agree

    3(d) agree with all reasons but independent is not accepted as it's given in question (saying it will give you no mark but you won't lose a mark for saying it as it's still right)
    Yeah 1b) is 0.669 just redone it, oh well I shouldn't have lost all 3 marks.
    3d) okay thank you.
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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?
    1)e) 3 marks

    3)b) 3 marks

    3)e)ii) 3 marks in total for part i) and ii). Definitely asked for SD but I got 16.6p

    My explanations about Gerald's claim were pretty much the same as yours, I just made a comment about the upper confidence limit in the first one as well. I'm glad you have mentioned the last part actually as I was unsure if I had done the right thing!
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    Could somebody please explain how you were suppose to go about 7b for me please.
    I used the rectangular distribution between 1 and 4 to calculate the E(X)=2.5 in that region, then integrated xf(x) between 4 and 6 and added 4 to it to calculate the E(X)=31/6 in that region. Then I averaged these two numbers to get somebody like 3.83. I didn't think I had got it right in the exam, but I'm curious about method marks for this question.
    I hope everybody else found the exam okay btw!:-)
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    (Original post by luke010203)
    Can you remember the island data from the table??
    For island B I think it was 116.9 and n was 7, so the mean was 16.7 but I can't remember what the standard deviation was to calculate the confidence interval


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    Yes that's right.

    You calculated s which is used as sigma (standard deviation) that was given in question as sigma(x-x bar) squared
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    (Original post by sam_97)
    1)e) 3 marks

    3)b) 3 marks

    3)e)ii) 3 marks in total for part i) and ii). Definitely asked for SD but I got 16.6p

    My explanations about Gerald's claim were pretty much the same as yours, I just made a comment about the upper confidence limit in the first one as well. I'm glad you have mentioned the last part actually as I was unsure if I had done the right thing!
    Thanks!

    So then I was right about my answer because everyone is saying it asked for variance? Makes me feel much better!

    I forgot about the first CI - did 18.7 lie in it because if so then I think I just got away with that haha!

    (Original post by benjammy)
    Could somebody please explain how you were suppose to go about 7b for me please.
    I used the rectangular distribution between 1 and 4 to calculate the E(X)=2.5 in that region, then integrated xf(x) between 4 and 6 and added 4 to it to calculate the E(X)=31/6 in that region. Then I averaged these two numbers to get somebody like 3.83. I didn't think I had got it right in the exam, but I'm curious about method marks for this question.
    I hope everybody else found the exam okay btw!:-)
    It was not a rectangular distribution, it was a cumulative distribution function - that's why!
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    (Original post by Suits101)
    Yes that's right.

    You calculated s which is used as sigma (standard deviation) that was given in question as sigma(x-x bar) squared
    Can you remember what sigma was because I got the wrong values for my confidence interval:/


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    Oh yh, i do remember doing sd for 3eii. Sorry about that to anyone who I told it was variance, i just remember the 27.77 but now remember doing the square root of that
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    (Original post by luke010203)
    Can you remember what sigma was because I got the wrong values for my confidence interval:/


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    Sorry I can't! I want to say something like 22.7 though.

    (Original post by MahuduElec)
    Oh yh, i do remember doing sd for 3eii. Sorry about that to anyone who I told it was variance, i just remember the 27.77 but now remember doing the square root of that
    Few! So it was 5.27 or something?
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    (Original post by luke010203)
    Can you remember what sigma was because I got the wrong values for my confidence interval:/

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    (x-xbar)^2 was around 22.6
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    (Original post by Suits101)
    Thanks!

    So then I was right about my answer because everyone is saying it asked for variance? Makes me feel much better!

    I forgot about the first CI - did 18.7 lie in it because if so then I think I just got away with that haha!



    It was not a rectangular distribution, it was a cumulative distribution function - that's why!
    But for 7a was the graph not rectangular between 1 and 4 at f(x)=0.25?
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    (Original post by Suits101)
    Sorry I can't! I want to say something like 22.7 though.



    Few! So it was 5.27 or something?
    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
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    (Original post by MahuduElec)
    (x-xbar)^2 was around 22.6
    I have no idea what I did wrong to get the values I did hopefully some method marks tho


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    (Original post by benjammy)
    But for 7a was the graph not rectangular between 1 and 4 at f(x)=0.25?
    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)
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    (Original post by luke010203)
    I have no idea what I did wrong to get the values I did hopefully some method marks tho


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    hopefully
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    (Original post by Suits101)
    Thanks!

    So then I was right about my answer because everyone is saying it asked for variance? Makes me feel much better!

    I forgot about the first CI - did 18.7 lie in it because if so then I think I just got away with that haha!
    It 100% asked for standard deviation

    I think it was 18.2 not 18.7, but yeah 18.2 lied inside the confidence interval
 
 
 
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