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December 1st, 2018, 09:16 AM   #1
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Median for grouped data

Hi,

I was studying a topic related to statistics and came across Median for Grouped data

Formula for finding it is pretty simple.
$\mathrm{Median} = L_m + \left [ \frac { \frac{n}{2} - F_{m-1} }{f_m} \right ] \times c$

where Lm is the lower limit of the median class, n is the total number of observations, Fm−1 is the cumulative frequency of the class preceding the median class, fm is the frequency of the median class, c is the class width.

But i wanted to know the derivation of this formula and came across https://math.stackexchange.com/quest...r-grouped-data and http://mathforum.org/library/drmath/view/71767.html which i found interesting.

I have some difficulty in finding the median of the example give at http://mathforum.org/library/drmath/view/71767.html

Quote:
 ----------------------------------------------------------- Class | Frequency | Cumulative Frequency ----------------------------------------------------------- 60-70 | 4 | 5 70-80 | 5 | 9 80-90 | 6 | 15 90-100 | 7 | 22 ----------------------------------------------------------- n=22 (even). So, i should take average of $\frac{n}{2}$ th and $\frac{n}{2} + 1$ th term? If the 6 values in the class 80 <= x < 90 are evenly spaced across these 10 units, then they are spaced 10/6 = 5/3 units apart. I would center them like this: This is where i am confused. I understand the data has in class has to be divided into 6 equal parts. But how can i find out which is the first data entry in this range (because we don't know the exact data entries)? Like here, Doctor Peterson has chosen 1st point to be 5/6 units farther than 80 (why?) Why couldn't it be some other point? 5/6 __5/3__ __5/3__ __5/3__ __5/3__ __5/3__ 5/6 / \ / \ / \ / \ / \ / \ / \ * * | * * * * +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+ 80 81 82 83 84 85 86 87 88 89 90 Therefore, the 2.5th value is 83 1/3 -- that is, 80 + 2*5/3, not 80 + 2.5*5/3. The standard formula gives Median = 80 + [(22/2) - 9] * (10/6) = 80 + 2*5/3 = 83 1/3
Sorry but i couldn't find a way to contact Doctor Peterson directly. Could you please help me out regarding this?

Thanks & Regards,
Dev

 December 5th, 2018, 09:50 AM #2 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 623 Thanks: 85 The problem is written wrong because multiple of ten are in two groups. For example, if one group ends at 80, the next group should start at >80, not 80. Alternatively, it could say something just above 80 like 80.01. Your division of 10/6 = 5/3 is correct. However, since 80 and 90 are in two groups, if the data points were 80, 81 2/3, 83 1/3, 85, 86 2/3, 88 1/3, and 90, it is impossible to say whether 5, 6, or 7 points go in the 80 to 90 range. I do not think it would be fair to mark any answer wrong because the problem is written wrong.

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