My Math Forum Median for grouped data

 Probability and Statistics Basic Probability and Statistics Math Forum

December 1st, 2018, 08:16 AM   #1
Newbie

Joined: Jun 2011

Posts: 17
Thanks: 0

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, Fmâˆ’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, 08:50 AM #2 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 639 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.

 Tags data, grouped, median

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post varuchi13 Advanced Statistics 2 March 30th, 2014 07:27 AM lumpa Real Analysis 0 October 19th, 2012 08:07 AM Alonso_Canada Advanced Statistics 3 March 29th, 2012 01:39 AM corwin43 Elementary Math 1 October 6th, 2009 04:18 PM BigLRIP Advanced Statistics 1 May 18th, 2009 10:01 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top