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    (Original post by iMacJack)
    Anyone offer some help with part F of this question please? I know this was in your paper Zacken - any help please? Thank you!
    You are told that c = Q3 + 1.5(Q3 - Q1)
    So find Q3 (part b) and Q1 (part a)
    Put the numbers into the equation.
    Similarly for d (d = Q1 - 1.5(Q3 - Q1)
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    (Original post by iMacJack)
    Anyone offer some help with part F of this question please? I know this was in your paper Zacken - any help please? Thank you!
    0.25 is the probability that exactly two candidates get a merit.

    So that 0.25^2. What's the probability that exactly 1 candidate passes but doesn't get a merit.

    Multiply that by 0.25^2 and then account for the # of ways this count happen.
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    (Original post by asinghj)
    You are told that c = Q3 + 1.5(Q3 - Q1)
    So find Q3 (part b) and Q1 (part a)
    Put the numbers into the equation.
    Similarly for d (d = Q1 - 1.5(Q3 - Q1)
    Person specifically asked for Q2(f).
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    What do you do in questions where the data changes and you have to show the effect on standard deviation, the correlation coefficient, Sxy... etc.

    https://1a388b28f5ce9318f837a64962fc...%20Edexcel.pdf

    Example here is 3f
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    For S1, there's types of questions where they state a number has been added or reduced. And then it asks you to state the effect made on the standard deviation, mean, IQR etc? Can someone please explain to me how to do these type of questions😩 I don't understand
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    (Original post by Zacken)
    0.25 is the probability that exactly two candidates get a merit.

    So that 0.25^2. What's the probability that exactly 1 candidate passes but doesn't get a merit.

    Multiply that by 0.25^2 and then account for the # of ways this count happen.
    Yeah I get that - I've got 3[0.25^2 * 0.75?] I can't see how it's (0.75-0.25) properly however? I don't quite understand the reasoning behind it
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    (Original post by Zacken)
    Person specifically asked for Q2(f).
    Oh yeah, he had part c highlighted so I thought that was part f 😂😂😂
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    Are we allowed to draw Normal and Standard Distribution diagrams in pencil for the exam?
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    (Original post by iMacJack)
    Yeah I get that - I've got 3[0.25^2 * 0.75?] I can't see how it's (0.75-0.25) properly however? I don't quite understand the reasoning behind it
    But 0.75 includes the probability that the third person gets a merit as well, in which case you'd get 3 merit marks when the question specifically wants exactly 2 merit marks. So the probability that the third person passes but does not get a merit is 0.75 - 0.25
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    (Original post by Philip-flop)
    Are we allowed to draw Normal and Standard Distribution diagrams in pencil for the exam?
    Yes.
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    (Original post by Zacken)
    But 0.75 includes the probability that the third person gets a merit as well, in which case you'd get 3 merit marks when the question specifically wants exactly 2 merit marks. So the probability that the third person passes but does not get a merit is 0.75 - 0.25
    But, why -0.25? Why not -0.01 or whatever? I'm being really stupid here I'm sure
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    Is p(a' n b') the same as p(a' u b')
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    (Original post by asinghj)
    January 2016 IAL was quite challenging.
    where did you get that paper from, i check edexcel website but login required ._.
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    (Original post by iMacJack)
    But, why -0.25? Why not -0.01 or whatever? I'm being really stupid here I'm sure
    75% pass the exam (this also includes getting a merit as they still pass)

    Getting a merit is 25% of students

    So just to pass and no merit is 0.75-0.25=0.5
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    (Original post by anndz3007)
    where did you get that paper from, i check edexcel website but login required ._.
    Here, Jan 16 attached
    Attached Images
  1. File Type: pdf S1.pdf (289.1 KB, 101 views)
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    (Original post by iMacJack)
    Here, Jan 16 attached
    Markscheme
    Attached Images
  2. File Type: pdf S1 Mark Scheme.pdf (514.3 KB, 65 views)
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    (Original post by iMacJack)
    Here, Jan 16 attached
    omg thank you so muchhh, gonna do it and june 15 tonight to prepare for the worst case scenario tmr lol
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    (Original post by metrize)
    75% pass the exam (this also includes getting a merit as they still pass)

    Getting a merit is 25% of students

    So just to pass and no merit is 0.75-0.25=0.5
    Okay but what I'm saying is even if they got 74% they'd still pass, but not get a merit, right? I'm a little bit confused...
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    (Original post by iMacJack)
    Okay but what I'm saying is even if they got 74% they'd still pass, but not get a merit, right? I'm a little bit confused...
    it say that 25% get merit, and i think you can only get merit once you passed, so those 25% is included in the 75%
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    (Original post by metrize)
    75% pass the exam (this also includes getting a merit as they still pass)

    Getting a merit is 25% of students

    So just to pass and no merit is 0.75-0.25=0.5
    I dont understand the logic behind this either
 
 
 
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