When a histogram is drawn from a table with unequal class widths, the bar heights must first be converted to frequency density. For finding the MEDIAN from the same unequal-width table:

Median = $l+\left(\frac{n/2-cf}{f}\right)\times h$ reads four things off the median class — its lower boundary $l$, its own frequency $f$, its own width $h$ — plus $n$ and the cumulative frequency of the class before it. Nothing is ever compared ACROSS classes, so unequal widths cause no distortion. A histogram is different: there the AREA of each bar must be proportional to its frequency, which is exactly why unequal widths force frequency density. Dividing the frequencies by the widths would wreck the cumulative count that $\frac{n}{2}$ is matched against, and merging classes to equal widths throws away information the table already gives.