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Can anyone help?
I know its the middle number, so Im just wondering whether I should use n/2 or (n+1)/2 to find the median? and is it the same for both listed and grouped data?
Thanks.
Last edited by Daniel123 (2007-05-24 05:10:43)
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median can not be found with a formula:
, the sum of all data, divided by the number of dataMode, is the data member that has the most occurence, so the mode of 1,1,2,3 is 1, the modes of 1,1,2,2,3 is 1 and 2
Median, is the middle number of an ordered list
so with the data 1,5,7,6,2,5,9 order them in ascending order: 1,2,5,5,6,7,9 so the median is 5
in the case of an even number of data entry's its the mean of the two middle numbers
1,5,7,6,2,9 -> 1,2,5,6,7,9 so the median is (5+6)/2 = 5.5
in terms of cumulative frequency, the median is the value taken at 50% up the dependant variable.
in terms of grouped data, draw a cumulative frequency table with the groups, and wherever the 50% hits, use that class as the median, and use the midpoint of the class.
Last edited by luca-deltodesco (2007-05-24 05:01:17)
The Beginning Of All Things To End.
The End Of All Things To Come.
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yes but what if there is an even number of values? wouldnt it be more sensible to add 1 before halving to find the 50% value? - and this could be done for both odd and even. Eg if there is 10 values, the middle value is the 5.5th - which would be (10+1)/2. If there is 9 values, the middle value would be the 5th - which would be (9+1)/2. I would just like to know when estimating the median from a c.f. diagram to make sure its as accurate as possible.
Last edited by Daniel123 (2007-05-24 05:23:18)
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!! i still dont have an answer i can use in the paper im doing right nowww!!!
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The way you're doing it is right. In an ordered list of n numbers, the [(n+1)/2]th number is the median.
Why did the vector cross the road?
It wanted to be normal.
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Thanks... is that the same for groups of numbers as well?
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I'm not sure what you mean, but I'll have a guess. Are you talking about something like:
Number | Frequency
-------------------------------
1 | 7
2 | 4
3 | 3
4 | 0
5 | 1
If so, then yes. You'd add up all the frequencies to get n (in this case, 15) and then use the same formula as before to get which number is the median. This time it would be the 8th number, which is a 2.
Why did the vector cross the road?
It wanted to be normal.
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Yep thats exactly what i meant! the papers ive been doing mostly ask to estimate the mean from a c.f. diagram... so i now know to use (n+1)/2 to find the middle value. Thanks
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