Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer

protected by Community May 1 '13 at 2:14

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.