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    (Original post by Delma)
    Argh for the IQR part of Q1 I used the 5.25th value for the LQ and the 15.25th value for the UQ so my IQR was 3.95. I don't know if that's right or wrong but it's close to 4 so I still got the outliers bit. I hope I don't lose too many marks for it though... can't believe the first question was the one I left until the end looool
    Out of the 7 marks you should only lose 1, assuming you did the test for part ii correctly. I think it was cruel to make the outlier boundaries so far away from the minimum and maximum values too as I must have checked the question 5 times before concluding I'd done it correctly the first time and moving on.
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    (Original post by jn998)
    I drew out the full tree diagram, took about 10 minutes to carefully check through the question and got 0.66. Probably wasted time doing that but I couldn't remember the method you were supposed to use
    A tree diagram was clever! Cause at least that way you remembered to multiply by 3 or whatever. I wrote out each possible combo i.e. WWL, WWD, DDW, DDL, LLW, LLD and then multiplied and added so I got 0.5^2 x 0.5 + 0.3^2 x 0.7 + 0.2^2 x 0.8. Ended up with 0.22 instead of 0.66 but should get majority of the marks.
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    How is the full exam paper on page 25?
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    (Original post by jn998)
    Out of the 7 marks you should only lose 1, assuming you did the test for part ii correctly. I think it was cruel to make the outlier boundaries so far away from the minimum and maximum values too as I must have checked the question 5 times before concluding I'd done it correctly the first time and moving on.
    Awesome, thanks. That seems about right. Pretty sure I did the test correctly, 1.5xIQR away from each quartile, right?
    Exactly! From all the past papers I've seen there's ALWAYS been at least one outlier but I'm sure they were just trying to catch us out this year haha.
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    Ahhhh what a beautiful S1 paper that was


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    What do you guys think the grade boundaries will be? I think it might be lower than last years


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    this is what i got:
    1i). For the stem and leaf diagram. Median = 29g
    Q1= 27.8 g
    Q2= 31.8g
    IQR = 4g
    ii) 1.5 x 4= 6
    31.8 + 6= 37.8g
    27.8-6= 21.8g
    So, no outliers.

    2i) The probability that team A does not lose is 0.5 * 0.5 * 0.5 = 0.125
    ii) 0.125 + (0.2)^3 +(0.3)^3=
    iii) 3(0.3*0.3*0.2) + 3(0.3*0.3*0.5) + 3(0.5*0.5*0.3) +3(0.5*0.5*0.2) etc you get the gist...

    3i) 5!= 120
    ii) 5P3= 60
    iii) 5C3= 10

    you round to 3 sig fig

    anyone wants to add to this so far?
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    this is what i got:
    1i). For the stem and leaf diagram. Median = 29g
    Q1= 27.8 g
    Q2= 31.8g
    IQR = 4g
    ii) 1.5 x 4= 6
    31.8 + 6= 37.8g
    27.8-6= 21.8g
    So, no outliers.

    2i) The probability that team A does not lose is 0.5 * 0.5 * 0.5 = 0.125
    ii) 0.125 + (0.2)^3 +(0.3)^3
    iii) 3(0.3*0.3*0.2) + 3(0.3*0.3*0.5) + 3(0.5*0.5*0.3) +3(0.5*0.5*0.2) etc you get the gist...

    3i) 5!= 120
    ii) 5P3= 60
    iii) 5C3= 10

    anyone wants to add to this so far?
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    (Original post by jaseho98)
    I think that you're right! Not gonna lie, a few of my friends thought the same as you. However, I just really find it stupid because it's not like we don't know how to do it... it's the damn phrasing sometimes which makes you interpret the question wrong. Then boom there goes all the marks :) damn

    Don't worry, I completely misinterpreted a question in C1 it's so easy to do XD
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    X~B( 20??? ,0.1)
    i) P(X=3) = 20C3 * (0.1)^3 * (0.9)^17
    ii) P(X less that or equal to 2)=
    use the tables
    iii) 0.1 * 20 =
    iv)Ho : p= 0.1 H1 : p<0.1v) there would be no critical region for that value of 20. 20 is too small. 12.16 % would have to be the significance levelvi) X~B (65,0.1)2 people use the pin 1234P(X less than or equal to 2)= 0.0531 > 5 percent Accept the Ho. There's insuffient evidence to suggest number of people using 1234 as PIN lowered.anyone want to add to this?
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    (Original post by chocolatecookieM)
    X~B( 20??? ,0.1)
    i) P(X=3) = 20C3 * (0.1)^3 * (0.9)^17
    ii) P(X less that or equal to 2)=
    use the tables
    iii) 0.1 * 20 =
    iv)Ho : p= 0.1 H1 : p<0.1v) there would be no critical region for that value of 20. 20 is too small. 12.16 % would have to be the significance levelvi) X~B (65,0.1)2 people use the pin 1234P(X less than or equal to 2)= 0.0531 > 5 percent Accept the Ho. There's insuffient evidence to suggest number of people using 1234 as PIN lowered.anyone want to add to this?
    No the significance level had to be 13% because it asked for the first integer level
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    3)iii) will not be 10. It was a probability so can't be an integer


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    2iii) I calculated it right now and it was 0.66.
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    (Original post by Mo2351)
    3)iii) will not be 10. It was a probability so can't be an integer


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    That sounds right, I think that's what I wrote
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    (Original post by chocolatecookieM)
    this is what i got:
    1i). For the stem and leaf diagram. Median = 29g
    Q1= 27.8 g
    Q2= 31.8g
    IQR = 4g
    ii) 1.5 x 4= 6
    31.8 + 6= 37.8g
    27.8-6= 21.8g
    So, no outliers.

    2i) The probability that team A does not lose is 0.5 * 0.5 * 0.5 = 0.125
    ii) 0.125 + (0.2)^3 +(0.3)^3
    iii) 3(0.3*0.3*0.2) + 3(0.3*0.3*0.5) + 3(0.5*0.5*0.3) +3(0.5*0.5*0.2) etc you get the gist...

    3i) 5!= 120
    ii) 5P3= 60
    iii) 5C3= 10

    anyone wants to add to this so far?
    For 2)i) The probability that team A doesn't lose is the probability that they draw or win (i.e. 0.5 + 0.3) and then there are 3 matches so its 0.8 * 0.8 * 0.8

    For 3)ii) There are 5! or 5P5 ways of getting a 5 letter combination as answered in part 1, but you're only asked for the probability of ABCDE or EDCBA. Because only 2 combinations satisfy what they're looking for it's 2/5P5 or 2/5! = 1/60.

    For 3)iii) Because the order matters in this case, there are 5P3 ways of getting a 3 letter combination. They only ask for ABC so because only this combination satisfies what they're looking for it's 1/5P3 = 1/60.

    Everything else looks alright to me. Correct me if I'm wrong though!
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    (Original post by GCSEsThen)
    No the significance level had to be 13% because it asked for the first integer level
    Can anyone confirm with this? so that's -1 mark.
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    (Original post by chocolatecookieM)
    Can anyone confirm with this? so that's -1 mark.
    Yeah sorry, it definitely asked for an integer value
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    (Original post by Delma)
    For 2)i) The probability that team A doesn't lose is the probability that they draw or win (i.e. 0.5 + 0.3) and then there are 3 matches so its 0.8 * 0.8 * 0.8

    For 3)ii) There are 5! or 5P5 ways of getting a 5 letter combination as answered in part 1, but you're only asked for the probability of ABCDE or EDCBA. Because only 2 combinations satisfy what they're looking for it's 2/5P5 or 2/5! = 1/60.

    For 3)iii) Because the order matters in this case, there are 5P3 ways of getting a 3 letter combination. They only ask for ABC so because only this combination satisfies what they're looking for it's 1/5P3 = 1/60.

    Everything else looks alright to me. Correct me if I'm wrong though!
    Yeah seems pretty right to me


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    Attachment 538319538327
    I found an almost identical question about the interquartile range on a stem and leaf diagram also with 20 values in the OCR (not MEI) June 2014 S1 paper (see attached) and in the mark scheme they accept loads of methods. It seems based on this that 4 is the main accepted answer, but 4.15 also would get full marks and a few other methods as well
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    Just some one markers i want to clarify
    Was it 0.5^3 or 0.8^3 for 2.i
    Was the reason for which we are unsure whether the mens or womens is more as a % because the data is grouped?
    And was k%, 13% ??
 
 
 
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