1
$\begingroup$

Given the formula for grouped median:

$Median = L_m + \left [ \frac { \frac{n}{2} - F_{m-1} }{f_m} \right ] \times c$

Where:

  • $L_m$: lower boundary of median class
  • $c$ : size of the median class
  • $F_{m-1}$ : cumulative frequency of the class before median class
  • $f_m$ : frequency of the median class
  • $n$ : size data

Example: What should the median be for the following:
- 100, 100, 100, 100, 100, 100, 100, 100, 100, 100 (a repeat of 100 ten times)?

Calculation:

Using a bin/class size of 0.5:

enter image description here

$L_m$ = 100

$c$ = 0.5

$F_{m-1}$ = 0*

$f_m$ = 10

$n$ = 10

100 + [(5-0)/10]*0.5

= 100.25

$\endgroup$
  • $\begingroup$ If a sample has ten $100$'s the median is not informative. Many books would say the median is $100.$ (Certainly the mean and the mode are 100.) // Your 'grouping' table does not make practical sense. But if you want to apply the formula to it, you could do that. // This is not a duplicate of any page I know of, but you might benefit from reading about getting the median from sensibly grouped data in a few of the pages under 'Related' in the right margin. $\endgroup$ – BruceET Jan 27 '18 at 3:07
1
$\begingroup$

When you group data into intervals, information is lost. So assumptions are made in order to make reasonable estimates of the sample mean, median, etc.

The assumption of this formula for estimating the median from grouped data is that the data are spread roughly uniformly throughout the interval. Clearly, this assumption is not met in your situation because all ten of the $100$'s lie at the lower endpoint of the interval. The idea of the formula is to estimate the median by interpolation, putting the estimate somewhere within the interval. In your case the estimated value $100.25$ is in the middle of the 'median interval' (the interval known to contain the median).

If you were trying to contrive a situation in which the estimate is even farther from the truth, you could put your ten $100$'s at the left end of an interval $[100, 120).$ With no other data, your estimate of the median would then be $110.$

There is nothing wrong with the formula, provided the assumption of data spread evenly throughout the interval is close to the truth. But any formula for estimating the median from grouped data will have to depend on assumptions. All that can be said for sure is the the median lies somewhere in the median interval. You have to recognize that the information lost in grouping data into intervals cannot be precisely recovered (unless the original data are saved and used).


Note: By contrast, the assumption usually made when trying to estimate the sample mean from grouped data is that each observation lies precisely at the midpoint of the interval that contains it. This idea gives rise to the formula $\bar X \approx \frac 1 n \sum_{i=1}^k f_jm_j,$ where there are $k$ intervals (usually of equal width), with midpoints $m_j$ and frequencies $f_j.$

$\endgroup$
  • 1
    $\begingroup$ This is helpful, I now appreciate that bin size and other assumptions will dictate the calculated grouped median. My intuition simply wasn't getting me there. $\endgroup$ – Minnow Feb 4 '18 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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