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In statistics, for grouped data, when calculating the median based on formula $Median = L_m + \left [ \frac { \frac{n}{2} - F_{m-1} }{f_m} \right ] \times c$

where $c$ is the size of the median class
$F_{m-1}$ is the cumulative frequency of the class before median class
$f_m$ is the frequency of the median class
$n$ is the total number of the data

I noticed some resources mentioned $L_m$ as lower class limit, but some lower class boundary. Which one is correct?

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What is the difference between "lower class limit" and "lower class boundary"? – Mike Spivey Sep 19 '11 at 4:31
If the class is something like 1-5, 6-10, 11-15, then lower class boundary is 0.5 for the first class, and lower class limit is 1. If the class is something like 0<x<=5, 5<x<=10, 10<x<=15, then lower class boundary for the first class is 0, and lower class limit is also 0. – Allen Sep 19 '11 at 4:39
That's helpful. Honestly, I'm not sure it really matters. The formula is an estimation of the median since we can't know it exactly; I've even seen $(n+1)/2$ instead of $n/2$ in the formula. See, for example, here. – Mike Spivey Sep 19 '11 at 5:05
Yes, I know that it is an estimation. Even there is some resources mentioned using the mid-point of the class. I just wonder why isn't there a standard explanation for the formula, or which one is the most reasonable, or which one is used in research. – Allen Sep 19 '11 at 7:04
Well, if you want an explanation you can see the answer I linked to in my previous comment. Because of the estimation aspect I really don't think there is one standard formula that's used everywhere. Getting back to your original question, then, I'm not sure that there is one that is "correct." – Mike Spivey Sep 19 '11 at 12:49
up vote 2 down vote accepted

After some consideration, in my opinion, "lower boundary" will make more sense rather than lower limit. For example, this is the data,

Class  Frequency
 1       1
 2       1
 3       1
 4       1

Based on the data, using we can know that the median is 2.5, without calculation. If using the formula as mentioned above, $\frac{n}{2}$ will get 2, there for the class contains the median is class 2, then using $L_m$ is a lower boundary,

$median = 1.5 + \left[ \frac{2 -1}{1}\right] \times 1 = 2.5$

This doesn't make sense for using lower limit. If changing the class to

Class Frequency
 1-2    1
 3-4    1
 5-6    1
 7-8    1

Using the method above, we will get,

$median = 2.5 + \left[ \frac{2 -1}{1}\right] \times 2 = 4.5$

However, if using class limit, then we will get 5.

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