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    (Original post by Nirm)
    When doing histograms and stuff. For calculating the width of the bars or for like area of the bars. Do you always go +/-0.5 of the two boundary numbers ( e.g. bar goes from 5 to 10 on the x-axis, so you take width to be 6 and not 5 as its 4.5-10.5). I am just confused by this (when do it) as in some questions you do it and some you don't.
    VV

    (Original post by hogree)
    If they use Values where Classes "share numbers" (so 60 is "in" the First Class and "in" the Second Class) like this:Then the Class Width is just 5. BUT if the Classes look like this...Then you use rounding.
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    (Original post by hogree)
    VV
    Ah okay that makes sense, thank you! Also anyone walk me through/roughly tell how you do part F and G of Question 7 on the June 2013 (R) Paper please? Do you always have to draw a table in questions like that or is there another (preferably shorter) method?
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    Don't know who it was - it was left unanswered, but to the person who asked - here you go

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    Good luck to everyone sitting this tomorrow! I shall create an opinion poll afterwards.
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    All the Best!!
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    Would appreciate it if someone could clarify how you get from step 1 to step 2 regrading y = a +bx.

    Step 1: f-100 = -0.424649..+ 0.395.. (m-100)
    Step 2: f= 0.735 +0.395m
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    When there's an outlier, what becomes the 'new' max/min value?
    Is it the the next value in the data, which isn't an outlier?
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    (Original post by Nirm)
    Ah okay that makes sense, thank you! Also anyone walk me through/roughly tell how you do part F and G of Question 7 on the June 2013 (R) Paper please? Do you always have to draw a table in questions like that or is there another (preferably shorter) method?
    The way I did it

    (f) Worked out the probability that S1 =1 which was 0.2
    Worked out the probability that S2 was 4 or less which was 0.8
    Multiply = 0.16

    And then g the way they do it in the mark scheme.

    (Original post by SANTR)
    Would appreciate it if someone could clarify how you get from step 1 to step 2 regrading y = a +bx.Step 1: f-100 = -0.424649..+ 0.395.. (m-100)Step 2: f= 0.735 +0.395m
    It's impossible if those are your values... What's the question?
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    (Original post by SANTR)
    When there's an outlier, what becomes the 'new' max/min value?
    Is it the the next value in the data, which isn't an outlier?
    yes
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    (Original post by hogree)
    The way I did it

    (f) Worked out the probability that S1 =1 which was 0.2
    Worked out the probability that S2 was 4 or less which was 0.8
    Multiply = 0.16

    And then g the way they do it in the mark scheme.

    It's impossible if those are your values... What's the question?
    Q3 part (b)

    http://qualifications.pearson.com/co..._June_2005.pdf
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    HEY can someone pls help me with june 2013(R) question 3 part e) I got 18.26 but I don't understand how to interpolate and get 12.. Here's the paper's link: http://qualifications.pearson.com/co...e_20130517.pdf
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    How do you do question 5b in this paper: https://0120b046f61e7ee5f95eb1315051...%20Edexcel.pdf

    AND........

    How do you do question 1 in this paper. I can't really picture how the tree diagram would look like
    https://0120b046f61e7ee5f95eb1315051...%20Edexcel.pdf
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    (Original post by wenogk)
    HEY can someone pls help me with june 2013(R) question 3 part e) I got 18.26 but I don't understand how to interpolate and get 12.. Here's the paper's link: http://qualifications.pearson.com/co...e_20130517.pdf
    I haven't done the question, but your 18.26 looks like a great number as you need to see how many plants are bigger than this. If you look at 18.26 you will see that it is roughly 1/3 along that class width (you can work it out exactly). That leaves you with 2/3 (=8 plants) + the 4 plants of the next class width make the total 12
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    (Original post by Florent venhari)
    How do you do question 5b in this paper: https://0120b046f61e7ee5f95eb1315051...%20Edexcel.pdf

    AND........

    How do you do question 1 in this paper. I can't really picture how the tree diagram would look like
    https://0120b046f61e7ee5f95eb1315051...%20Edexcel.pdf
    Question 5b is really asking P( 20< T < 30). Does that make sense?
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    (Original post by candol)
    I haven't done the question, but your 18.26 looks like a great number as you need to see how many plants are bigger than this. If you look at 18.26 you will see that it is roughly 1/3 along that class width (you can work it out exactly). That leaves you with 2/3 (=8 plants) + the 4 plants of the next class width make the total 12
    I really don't get that, can you tell me what the class width is - is it 5 because its in the 5< y <10 class... Would you be kind enough to do this question plsss
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    You plugged in m-100 rather than m-250
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    If in my working out the correct answer is shown, then afterwards i put something incorrect (such as the answer being 0.9772 but afterwards i did 1-0.9772=0.0228) would they ignore the incorrect or penalise me?
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    can someone please explain to me why theres different formulas for Variance and when you'd use the different ones?
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    (Original post by wenogk)
    HEY can someone pls help me with june 2013(R) question 3 part e) I got 18.26 but I don't understand how to interpolate and get 12.. Here's the paper's link: http://qualifications.pearson.com/co...e_20130517.pdf
    Here's how I did it:
    1 standard deviation above the mean has a yield of 18.3 so you need to look at the group 15<y<25
    So for normal interpolation you do
    Lower bound + how far in group/frequency of group *class width =18.3
    and so rearrange to get how far in the group=18.3 - LB /class width *frequency

    So
    15 + ?/12 * 10=18.3
    rearrange ?=3.96
    But this shows the number of plants that have a yield between 15<y<18.3
    and you want above so in that group there are 12 plants so 12 - 3.96 + the extra 4 in the next group =12.04 so 12 plants.
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    (Original post by haes)
    If in my working out the correct answer is shown, then afterwards i put something incorrect (such as the answer being 0.9772 but afterwards i did 1-0.9772=0.0228) would they ignore the incorrect or penalise me?
    As long as you didn't cross it out, you'd get the marks
 
 
 
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