Solved – (Frequency Histogram) – Bins of equal width

It is not exactly clear whether this is an elementary question which you should be able to tackle on your own, but I'll address it.

The first step is to summarize your data. For your example I get minimum $0.15$ and maximum $0.98$.

Bins are best chosen with nice numbers in mind. "Nice numbers" are a little hard to define but easy to recognise. They are based on powers of $10$ and multiples of $1$, $2$ and $5$.

Unless your data start well above $0$, or well below $0$, you should always consider including $0$ as a bin boundary.

Here a range from $0$ to $1$ seems obvious, at least with some experience. $5$ bins would mean intervals of $0.2$; $10$ of $0.1$ and $20$ of $0.05$.

In choosing bins it is vital to be careful, consistent and explicit about what happens if a value falls on the edge of a bin. One set of choices would be

• $0$ or more but less than $0.2$
• $0.2$ or more but less than $0.4$

and so forth. So the lower limit of each bin is included. The uppermost bin can be different. A common notation for that would be $[0, 0.2)$ and so forth.

For different measurement schemes, some flexibility is needed. Thus for degrees on a circle, bins of width $90^circ$ with limits $[0, 90), [90, 180), [180, 270), [270, 360)$ would be much, much better than bins of width $100$.