Does the median tend to the lower limit of the median class? The median for a continous distribution is given by 
$M= L +\dfrac{(N/2-C)\cdot I}{f}$
where $M$ is median, $L$ is lower limit of median class, $N$ is the total frequency, $C$  is the cumulative frequency of class BELOW the median class, $F$ is frequency of median class, $I$ is class width. The way I see it $N/2$ is close to the medial value. $(N/2-C)$ is hence the frequency of the values in the medial class lesser than the median. You then express that as a fraction of the total class size and add it to the lower limit.
So, if there is a higher percentage of values in the $N/2$ to $C$ range, the median should tend to the lower limit of the medial, class, right?
But then, in this case, $(N/2-C)$ is high, so the median will tend to the upper limit, which is not right.
Where is the mistake?
Thanks in advance
 A: The possibility that there is a higher percentage of values in the $C$ to $N/2$ range   does not imply that the median tends to the lower limit of the median class. This formula is based on an interpolation that assumes a uniform distribution of data within the median class. Consider the fraction $\displaystyle\frac {N/2 - C}{f}$ (where $C$ is the cumulative frequency of the interval below the median class and $f$ is the frequency of the median class), which represents the proportion of data values in the median class that are below the median. 
 If we assume that data are uniformly distributed (i.e., equally spaced) within the median class, we can multiply the fraction by $I$ (the width of the whole median class) to get the fraction of median class width corresponding to the position of the median. Adding the result to $L$, we get the final formula of the median. 
The fact that a larger number of data in the lower portion of the median class do not move the median towards lower values, but rather towards values, is therefore the consequence of the assumption of uniform distribution within the median class.
