Why would you use a different calculation for finding median from a histogram rather than from a regular set of data? I've always learned that you should use $\frac{n+1}{2}$ to find the median of a regular set of data.  For example, the median of $$1, 2, 3, 4, 5, 6, 7$$ would be $4$ because of $\frac{n+1}{2}$.
But question #3 on this website about histograms uses $\frac{n}{2}$ in the answers.  Why would you use $\frac{n}{2}$ to find the median from a histogram, but then use $\frac{n+1}{2}$ for a regular set of data?
 A: The difference you're noticing is mainly because of the difference between even and odd $n$. But histograms are in general trickier to deal with, because there is less information, so let me go through both issues.
Suppose you're given all the numbers in a sequence like $1, 2, 5, 10, 20, 50, 100$. After sorting it, you see that there are $n=7$ elements, so you take $\frac{n+1}{2} = 4$ and the median is the value of the $4^{\text{th}}$ element: $10$.
Suppose the sequence has $n=6$ elements instead: $1, 2, 5, 10, 20, 50$. Then $\frac{n+1}{2} = 3.5$ and we average the $3^{\text{rd}}$ and $4^{\text{th}}$ elements to get $7.5$ as the median. Note that in this case, the median is a value with $\frac n2$ of the elements on one side, and $\frac n2$ elements on the other side! ($3$ of the numbers are below $7.5$, and $3$ are above $7.5$.)

Additionally, when you're given a histogram, you have less to work with. In an example like yours, we know that there are $n=60$ elements, and they're in the following bins:

*

*$9$ of them are between $60$ and $75$.

*The next $16$ are between $75$ and $85$.

*The next $25$ are between $85$ and $90$.

*The next $10$ are between $90$ and $110$.

Since $\frac{n+1}{2} = 30.5$, we are looking for the value halfway between the $30^{\text{th}}$ and the $31^{\text{th}}$ element. Another way to say it: $30$ of the numbers are below the median, and $30$ of the numbers are above the median. Both of these things are exactly the same thing we'd say if we had the whole sequence in front of us!
The only extra challenge with the histogram is that we don't actually know what the $30^{\text{th}}$ element or the $31^{\text{th}}$ element are. We know that the range $85$ to $90$ contains the $26^{\text{th}}$ through $50^{\text{th}}$ element. So both the $30^{\text{th}}$ and the $31^{\text{th}}$ elements are roughly $1/5$ of the way through this range: they are both about $86$. This means that $86$ is also a good guess for the midpoint between the two.
(It's possible that we're wrong. For example, maybe all $25$ elements in the $85$-to-$90$ range are equal to $89$ exactly. Then the true median will also be $89$, and our guess is wrong.)
In the case of a histogram:

*

*All we can do is estimate.

*The value of $n$ is likely to be very large.

Both of these mean that it isn't really worth it to care about the difference between $\frac n2$ and $\frac{n+1}{2}$. But it doesn't mean we should do anything different for a histogram that we don't do for a sequence.
