Basic transistor audio amplifier bias question
As far as I understand your problem, the key point is in the sentence (quote):  "In order to provide a stiff base voltage, resistor R4 should have a current of about 5 to 10 times the base current. For this example, we will assume 9 times the base current for a total of .9 ma."

In general, the voltage divider is chosen with the aim to produce a base voltage Vb "as stiff as possible". That means: As independent as possible on the base current which is connected with large tolerances (due to B variations). (Due to some other constraints like power consumption and input resistance the voltage divider must not be much lower in resistance. Without these constraints, it would be even better to chose a factor between both currents of 20 or 50 instead of only 9).

This is the typical design approach because - at the same time - the emitter resistor R2 provides current controlled voltage feedback (voltage Ve) which can work satisfactorily only for a "stiff" base voltage Vb. As a consequence of the negative feedback, the collector current does not depend too much from the base-emitter voltage, which normally is chosen (for calculating purposes) to be app. Vbe=0.65 volts.  

It is very easy to visualize this effect using the method of "stabilizing line" in the Ic=f(Vbe) characteristic. This is nothing else than to plot the ohmic law for the resistor R2 - expressed not by the emitter voltage Ve but by Ve=Vb-Vbe. In this case, both graphs can be plotted in one single diagram - and it becomes clear that Ic does depend only very little on the temperature-dependent Ic=f(Vbe) characteristic. The actual DC operating point exist where both curves cross each other - the exponential BJT characteristic and the linear R2 characteristic Ic=f(Vb-Vbe), which has a negative slope.

The enclosed drawing shows the construction of the stabilization line.