Median estimated from grouped data with a single class 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:

$L_m$ = 100
$c$ = 0.5
$F_{m-1}$ = 0*
$f_m$ = 10 
$n$ = 10
100 + [(5-0)/10]*0.5 
= 100.25

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 $f_j.$
