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    I really don't understand what to do for part C, this is from the June 2014 R paper (edexcel), would appreciate some help.

    Thanks.
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    (Original post by DeRerumNatura)
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    I really don't understand what to do for part C, this is from the June 2014 R paper (edexcel), would appreciate some help.

    Thanks.
    The height is frequency density = frequency / class width (full video tutorial here).

    For part C is asking you to use linear interpolation, which you would first do by identifying at which class intervals the mean lies. So firstly, do a cumulative frequency table and use n/2 to find the frequency of the mean. Then figure at which group it belongs to and use this simple method:
    lower boundary of class + class width x (group frequency - mean frequency / group frequency)
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    (Original post by kkboyk)
    The height is frequency density = frequency / class width (full video tutorial here).

    For part C is asking you to use linear interpolation, which you would first do by identifying at which class intervals the mean lies. So firstly, do a cumulative frequency table and use n/2 to find the frequency of the mean. Then figure at which group it belongs to and use this simple method:
    lower boundary of class + class width x (group frequency - mean frequency / group frequency)
    Hi,

    Thanks, but how would I do this to get the number of people waiting between 3.5 and 7, which is what the question asks ? And why would you do n/2 I thought that was to find the median ?

    Thanks
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    (Original post by kkboyk)
    The height is frequency density = frequency / class width (full video tutorial here).

    For part C is asking you to use linear interpolation, which you would first do by identifying at which class intervals the mean lies. So firstly, do a cumulative frequency table and use n/2 to find the frequency of the mean. Then figure at which group it belongs to and use this simple method:
    lower boundary of class + class width x (group frequency - mean frequency / group frequency)
    This is what they do in the mark scheme :

    1/3*15 + 9 + 1/2*6 = 17
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    (Original post by DeRerumNatura)
    Hi,

    Thanks, but how would I do this to get the number of people waiting between 3.5 and 7, which is what the question asks ? And why would you do n/2 I thought that was to find the median ?

    Thanks
    That was for part d, where you use n/2 for the median.

    For C you should use your histogram, since for the previous part you were told to find the height of the group 2-4. Then find the area of the bar from between 3.5 - 4, for 5-6 (which is just 9 from the table) and for 7 (which is just 6 from the the table)

    (Original post by DeRerumNatura)
    This is what they do in the mark scheme :

    1/3*15 + 9 + 1/2*6 = 17
    The bit in bold above
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    (Original post by kkboyk)
    That was for part d, where you use n/2 for the median.

    For C you should use your histogram, since for the previous part you were told to find the height of the group 2-4. Then find the area of the bar from between 3.5 - 4, for 5-6 (which is just 9 from the table) and for 7 (which is just 6 from the the table)



    The bit in bold above

    Hi,

    Thanks so much, I understand it now, I didn't think to use the histogram.
 
 
 
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