A regulator has an input of 15 V, a series resistor of 120 Ω and a 6.2 V Zener diode…

Step 1. The series current. $I_s = \dfrac{15\ \text{V} - 6.2\ \text{V}}{120\ \Omega} = 73.3$ mA.

Step 2. The greatest load current. The Zener diode must keep $4.0$ mA, so $I_{L,max} = 73.3\ \text{mA} - 4.0\ \text{mA} = 69.3$ mA.

Step 3. The smallest load. $R_{L,min} = \dfrac{V_z}{I_{L,max}} = \dfrac{6.2\ \text{V}}{0.0693\ \text{A}} = 89.4\ \Omega$.

Step 4. A load of $60\ \Omega$. This is below $89.4\ \Omega$, so the Zener diode is off. The series resistor and the load divide the input: $V_{out} = 15\ \text{V} \times \dfrac{60\ \Omega}{120\ \Omega + 60\ \Omega} = 5.00$ V.