In the canonical ensemble the system's energy is not fixed. So what values does it actually take? Two things decide that.
One is the number of states. The higher the energy the more violently they multiply. That is the rising curve in the figure. The other is the Boltzmann factor, which shrinks exponentially as the energy rises. That is the falling curve.
The energy we actually see goes as the product of the two. Multiply the rising side by the falling side and you get the peak in the lower part of the figure. The system's energy is not fixed, and yet it hardly ever leaves that peak.
What happens to the width of the peak as the particles multiply? The answer first: the width seen as a fraction of the mean shrinks as \dfrac{1}{\sqrt{N}}. Exactly the form the very first piece had.
The figure takes the horizontal axis divided by the mean. Add particles and the peak clings ever closer to that mean.
The reason is the first piece's reason too. The energy is the sum of what N particles hold, and the more terms are added, the closer the total gathers about its mean. At 10^{23} an energy that was supposedly undetermined becomes, for all purposes, a single value.
To put a number to the spread of the peak we use the variance, \langle E^2 \rangle - \langle E \rangle^2, whose square root is the width drawn in the figure.
This too comes out of the partition function. The previous piece got the mean from one derivative of \log Z with respect to \beta. A second derivative gives the variance: \langle E^2 \rangle - \langle E \rangle^2 = \dfrac{\partial^2 \log Z}{\partial \beta^2}.
The first derivative is the centre, the second is the spread. Simply differentiating the partition function again and again gives the shape of the distribution in turn.
Rewrite that variance from the side of temperature. Turn the derivative in \beta into one in T and it becomes \langle E^2 \rangle - \langle E \rangle^2 = kT^2 C, where C is the heat capacity, the heat needed to raise the temperature by 1.
The left side is how widely the system wanders when left alone. The right side is how it answers heat given from outside. Two things of quite different character are being called the same quantity.
The figure has the curve of \langle E \rangle against temperature with a band of fluctuation about it. Where the curve rises steeply, the band thickens too. A system that wanders freely takes in heat freely. This is the simplest instance of the relations tying fluctuation to response.
Two ensembles have now appeared: the microcanonical with the energy fixed, and the canonical with the temperature fixed. Which of them is right?
The figure lays the canonical distribution of energy over the single value the microcanonical fixes. As the particles multiply, the distribution clings to that one line.
So the answer is that it makes no difference. With N large the difference is of order \dfrac{1}{\sqrt{N}} and cannot be measured. One may therefore take whichever is easier to solve. Usually that is the canonical, and the next chapter keeps to it.