Substituting into Equations 12.20 and 12.21, we find that

R_B=\frac{1}{1/R_1+1/R_2} =3.33 \text{ k}\Omega \\ V_B=V_{CC}\frac{R_2}{R_1+R_2}=5 \text{ V}

Then, substituting into Equation 12.23 and using β = 100, we have

I_B=\frac{V_B-V_{BE}}{R_B+(\beta +1)R_E}=41.2 \ \mu \text{A}

For β = 300, we find that I_B = 14.1 \ \mu\text{A}. Notice that the base current is significantly smaller for the higher β.
Now, we can compute the collector current by using I_C = \beta I_B. For β = 100, we find that I_C = 4.12 \text{ mA}, and for β = 300, we have I_C = 4.24 \text{ mA}. For a 3:1 change in β, the collector current changes by less than 3 percent. The emitter current is given by I_E = I_C + I_B. The results are I_E = 4.16 \text{ mA} for β = 100 and I_E = 4.25 \text{ mA} for β = 300.
Finally, Equation 12.24 can be used to find V_{CE}. The results are V_{CE} = 6.72 for β = 100 and V_{CE} = 6.51 for β = 300.