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Grouped Frequency Table Median?

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    Hey guys,

    Just wanted to ask a quick question, How do you work out the median of a Grouped frequency table? I heard you use formula which is n+1/2 and what is n?.....Also one you have found out the answer how do you know which group it's in?
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    n I think is your total frequency then once you've found the group its in I'm pretty sure you interpolate
    equation b-((n/2)-f)/(fm)xw
    where b is lower class bound
    f is sum of frequencies below class
    fm is frequency of median class
    w is class width
  3. Offline

    ReputationRep:
    (Original post by Nelo Angelo)
    n I think is your total frequency then once you've found the group its in I'm pretty sure you interpolate
    equation b-((n/2)-f)/(fm)xw
    where b is lower class bound
    f is sum of frequencies below class
    fm is frequency of median class
    w is class width
    Thanks but i'm not talking about upper bounds and lower bounds, and i'm not talking about histograms:confused:
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    So nobody knows on TSR how to find out the median on a grouped frequency table?
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    one of the guys above said it - you interpolate.

    If its grouped then you're assuming the data is continuous so you have to interpolate;

    m + cw[(0.5n-fc)/f]

    where m = minimum boundary of median class
    cw = class width of median class
    n = total number in the sample
    fc = cumulative frequency of the classes below the class
    f = frequency of the median class
  6. Offline

    ReputationRep:
    1-n+1/2=position of the median
    n=number of values or the sample size
    2-you then need to add a cumlative frequency column to the table and the number you got for 1- needs to be fitted in. The group it is in is the median group, but i'm not sure how you work out the actual value.

    for example-
    weight frequency cumlative frequency
    1<n<2 4 4
    2<n<3 10 14

    n+1/2=14+1/2=7.5th number
    7.5 is larger than 4, but less than 14, so is in the 2<n<3 group
    2<n<3 is the median group.

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