Finding $Q_1, Q_3$ for a set of data. The Problem I'm Given:
Suppose the following data are obtained by recording $X$, the number of customers that arrive at an automatic banking machine during $15$ successive one-minute
time intervals.

Record the median and IQR and deside whether there are any outliers. 

To start, I put them all from lowest to greatest
$$0,0,0,0,1,1,1,2,2,2,2,3,3,4,4 $$
Median would be the middle value being, since its odd, $2$ ($(15+1)/2 = 8$)
How do I find $Q1$ and $Q3$?
 A: $Q_1$ is essentially the median of the data that are less than the overall median, and $Q_3$ the median of the data greater than the overall median.
This is a little nuanced so I'll elaborate a bit.
First, I feel like it will help to write your data into two sets, because we have multiple $2$'s that can muck up my explanation. We divide it up at the median:
$$0,0,0,0,1,1,1,\;\;\;\;\;(2),\;\;\;\;\;2,2,2,3,3,4,4$$
Note that if you have an even number of elements in the data set (the case where you have to find the average of the middle two values) you would just divide the set into two equal halves. (For example if we omitted the "circled" two from the above data set, the set on the left and the right would be just as they are now, we just wouldn't have anything in the middle.)
Now, I "circled" $2$ specifically because we have an odd number of numbers in the data set. We have to handle these sorts of cases differently. Here, we put the median ($2$) into both the upper and lower sets now:
$$0,0,0,0,1,1,1,(2),\;\;\;\;\;\;\;\;\;\;(2),2,2,2,3,3,4,4$$
The median of the left set is $Q_1$ and the median of the right side is $Q_3$.

If we had an even number of data points instead, and thus nothing was circled, we wouldn't add anything into either half, we would just divide them into two and find the median of each half. For example, if our data was
$$1,2,3,4,5,6,7,8,9,10$$
the median is $(5+6)/2 = 5.5$. Since we had an even number of data points, we just divide into two halves: the lower half that's below the median, and the upper half that's above it
$$1,2,3,4,5,\;\;\;\;\;6,7,8,9,10$$
Here, then, $Q_1$ is the median of the lower set ($3$), and $Q_3$ the median of the upper set ($8$).

Edit: I couldn't remember the exact convention for how people find $Q_1,Q_3$, looked it up, saw I was wrong. I deleted my post, corrected it, and then reposted it.
