Binning By Equal-Width I have a dataset:
5, 10, 11, 13, 15, 35, 50 ,55, 72, 92, 204, 215

The formula for binning into equal-widths is this (as far as I know)
$$width = (max - min) / N$$
I think N is a number that divides the length of the list nicely. So in this case it is 3.
Therefore:
width = 70
How do I use that 70 to make the bins?
 A: \begin{align}\text{width}= \frac{\max-\min}{ \text{number of bins}}
    = \frac{215-5}3
    =70.
\end{align}
$$5, 10, 11, 13, 15, 35, 50 ,55, 72, 92, 204, 215$$

*

*$70+5=75$  (from $5$ to $75)=$Bin $1: 5,10,11,13,15,35,50,55,72$

*$70+75=145$ (from $75$ to $145)=$Bin $2:92$

*$70+145=215$ (from $145$ to $215)=$Bin $3: 204,215 $
A: So far, all the answers have proposed a representation of the histogram that is, in a sense, biased.  Bias is not necessarily bad, but it is good to recognize it and to be able to control it.
The bias arises because these methods establish a set of bins in which the smallest data value is at the extreme left of the lowest bin and the largest data value is at the extreme right of the highest bin.  The bin width is thus the smallest possible that could be used to stuff these data into a desired number of bins.  (For some purposes that is ideal.)  A more representative bin width would be one that looked as if the bins had not been chosen on the basis of the data.  That's more useful for evaluating the histogram in any context where the data are viewed as a random sample.  In such a circumstance there would, on average, be about a half a bin occupied beyond each of the data extremes.
For a less biased representation, divide the range of the data $(x_i)$ into $n-1$ pieces rather than $n$ and situate the bins so that the outer halves of the extreme bins are empty.  More generally, if you would like to effect some compromise between these approaches, choose a positive number $a$ strictly between $0$ and $2$ and divide the range into $n-a$ pieces.  This can immediately be done with the ceiling transformation

$$x \to \lceil \frac{x - m + \delta}{h} \rceil$$

where $m$ is the least value in the data, $M$ is the greatest, the bin width is
$$h = \frac{M-m}{n-a},$$
and
$$\delta = \frac{nh - (M-m)}{2h}$$
is the amount of extra space on either side of $m$ and $M$ in the histogram.
(BTW, if you wish to specify the bin width $h$ instead of the number of bins, you may independently set the value of $\delta$ (representing the empty space at the left of the leftmost bin) and simply apply the first formula.  It will give an integer -- positive when $\delta$ is positive -- indicating the bin assignment.)
When $a=0,$ the minimum data values will be put into a nonexistent bin $0.$  It is conventional to assign them to bin $1.$   (An easy way to accomplish this without an extra programming step is to set $a$ to a very small value, such as $10^{-6}.$)  As $a$ increases, the extreme bins become less and less occupied
To illustrate, I generated $50$ values from a uniform distribution and computed $n=4$ equally spaced bins for a range of values of $a.$

The gray backgrounds show the extent of each bin while the heights of the horizontal segments (which are always at the same locations in each plot, since the data never change) depict the data values.  They are colored according to their bin assignments.  You can see how with $a=0$ the data fully fill the bins while with $a=2$ only the smallest and greatest data values occupy the extreme bins.
Here is the R code I used for assigning arbitrary values y to the bins determined by the data x.
bin <- function(y, x, n, a=1) {
  if (missing(x)) x <- y
  r <- range(x, na.rm=TRUE)
  h <- diff(r) / (n - a)
  delta <- (n * h - diff(r)) / 2
  ceiling((y - r[1] + delta) / h) 
}

A: I found the answer. I was somewhat close with the question. The trick is that width is not just width, it is width of each interval. 
Therefore
bin1: 5,10,11,13,15,35,50,55,72 I.e. all values between 5 and 75
bin2: 92 I.e. all values between 75 and 145    
bin3: 204,215 I.e. all values between 145 and 215

And for equal width binning you are given number of required bins and in this case it is 3.
A: Binning is a unsupervised technique of converting Numerical data to categorical data but it do not use the class information. There are two unsupervised technique.
1-Equal width.
2-Equal frequency.
In Equal width, we divide the data in equal widths. In order to calculate width we have the formula.
width=(max−min)/N
So from the available data in the problem we have 75 is the width.
We divide the data with three categories:
5 to 75
75 to 145
145 to 215.
Now we keep the numbers in the above ranges.Thus,
bin1: 5,10,11,13,15,35,50,55,72
bin2: 92 will hold only one value as we have only one value exist between 75 and 145
bin3: 204,215.
A: I also stumbled on the problem about how to exactly divide the bins, and could not find anything on the internet. So here is how I would formalize the problem to unambiguously assign the items to the $N$ desired bins.
Provided some multiset $L \in \mathbb{R}^n$, $n \in \mathbb{N}^+$ and the number of desired bins $N \in \mathbb{N}^+$.
Calculate the width $w \in \mathbb{R}$ like so (note that $w$ does not have to be a natural number):
\begin{equation*}
w := \frac{1}{N} \cdot \bigl(\max(L)-\min(L)\bigr)
\end{equation*}
Then calculate the intervals $I_1, \dots, I_N$:
\begin{equation*}
I_i :=
\begin{cases}
\Bigl[\min(L) + (i - 1) \cdot w,\ \min(L) + i \cdot w\Bigr),\ &i \in \{1, \dots, N-1\}\\
\Bigl[\min(L) + (N - 1) \cdot w,\ \min(L) + N \cdot w\Bigr], &i = N
\end{cases}
\end{equation*}
Alternatively, you can use:
\begin{equation*}
I_i :=
\begin{cases}
\Bigl[\min(L),\ \min(L) + w\Bigr], &i = 1\\
\Bigl(\min(L) + (i - 1) \cdot w,\ \min(L) + i \cdot w\Bigr],\ &i \in \{2, \dots, N\}
\end{cases}
\end{equation*}
Then fill each bin $B_1, \dots, B_N$ like so:
\begin{equation*}
B_i := \bigl\{x \in L \mid x \in I_i\bigr\} = L \cap I_i,\ i \in \bigl\{1, \dots, N\bigr\}
\end{equation*}
Pseudocode for inputs L[0, ..., n-1] and N desired bins, choosing the last interval as largest:
L := sorted(L)
width := (1/N) * (L[n-1] - L[0])
B[i] := [] for i in {0, ..., N-1}
i := 0
for x in L:
    if x >= L[0] + i * width and i != N-1:
        i := i + 1
    B[i] := B[i] + [x]

With your example $L := \{5, 10, 11, 13, 15, 35, 50 ,55, 72, 92, 204, 215\}$ and $N := 3$, we get the following:
\begin{equation*}
w = \frac{1}{N} \cdot \bigl(\max(L) - \min(L)\bigr) = \frac{1}{3} \cdot (215 - 5) = 70\\
\begin{aligned}
I_1 &= \left[5, 75\right) \longrightarrow B_1 = \{5,10,11,13,15,35,50,55,72\}\\
I_2 &= \left[75, 145\right) \longrightarrow B_2 = \{92,204\}\\
I_3 &= \left[145, 215\right] \longrightarrow B_3 = \{215\}
\end{aligned}
\end{equation*}
