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    Hey guys
    S1 is just around the corner now and I've been doing past papers and I always see questions that say 'use linear interpolation formula to find the median' and its usually for 2 marks. I know the comparison method but I really dont like it and takes too much time, is there an alternate way ? maybe some sort of formula ?


    Thanks
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    start of group + [(amount that median is into group)/frequency] * group width
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    (Original post by Lunch_Box)
    start of group + [(amount that median is into group)/frequency] * group width
    Amount of median ?
    I saw this b + <(0.5n - f)/ f > * C
    but dont know what the letters represent and what 0.5 is
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    Median is Q2 so do n/2 and round up if it's not a whole number. (Where n = total freq) Then use interpolation.

    Let's use some example data.

    class | frequency
    10 < a < 20 | 20
    20 < a < 30 | 30
    30 < a < 40 | 40

    Total freq = n = 20+30+40 = 70

    n/2 = 35

    We are looking for the 35th value. This value clearly lies within 20 < a < 30 and we know we are looking for the (35-20 =) 15th value into this class.

    Calculate the class width. In this case it is 30-20 = 10. (if the classes were 10-19, 20-29 then the class width is 29.5-19.5)

    Now do class width/freq = 10/30 = 1/3

    Multiply this by the number into the class.

    1/3 * 15 = 15/3 = 5

    Now do lower bound + 5 = 20+5 = 25.

    This leaves us with: Lower bound of class + [class width/frequency] * number into the class
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    (Original post by JackClarke)
    Median is Q2 so do n/2 and round up if it's not a whole number. (Where n = total freq) Then use interpolation.

    Let's use some example data.

    class | frequency
    10 < a < 20 | 20
    20 < a < 30 | 30
    30 < a < 40 | 40

    Total freq = n = 20+30+40 = 70

    n/2 = 35

    We are looking for the 35th value. This value clearly lies within 20 < a < 30 and we know we are looking for the (35-20 =) 15th value into this class.

    Calculate the class width. In this case it is 30-20 = 10. (if the classes were 10-19, 20-29 then the class width is 29.5-19.5)

    Now do class width/freq = 10/30 = 1/3

    Multiply this by the number into the class.

    1/3 * 15 = 15/3 = 5

    Now do lower bound + 5 = 20+5 = 25.

    This leaves us with: Lower bound of class + [class width/frequency] * number into the class
    Thank you
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    (Original post by JackClarke)
    Median is Q2 so do n/2 and round up if it's not a whole number. (Where n = total freq) Then use interpolation.

    Let's use some example data.

    class | frequency
    10 < a < 20 | 20
    20 < a < 30 | 30
    30 < a < 40 | 40

    Total freq = n = 20+30+40 = 70

    n/2 = 35

    We are looking for the 35th value. This value clearly lies within 20 < a < 30 and we know we are looking for the (35-20 =) 15th value into this class.

    Calculate the class width. In this case it is 30-20 = 10. (if the classes were 10-19, 20-29 then the class width is 29.5-19.5)

    Now do class width/freq = 10/30 = 1/3

    Multiply this by the number into the class.

    1/3 * 15 = 15/3 = 5

    Now do lower bound + 5 = 20+5 = 25.

    This leaves us with: Lower bound of class + [class width/frequency] * number into the class
    Are there any videos on youtube which may help me understand it better ?
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    Estimated Median = L + (n/2) − B × w
    G
    where:

    L is the lower class boundary of the group containing the median
    n is the total number of values
    B is the cumulative frequency of the groups before the median group
    G is the frequency of the median group
    w is the group width
 
 
 
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