The compound shown is 2-bromo-3-methylpent-2-ene drawn with a definite geometry. Its configuration is

On C2 the groups are $\mathrm{Br}$ and $\mathrm{CH}{}_3$; $\mathrm{Br}$ has priority. On C3 the groups are $\mathrm{C}{}_2\mathrm{H}{}_5$ and $\mathrm{CH}{}_3$; ethyl has priority. In the drawing $\mathrm{Br}$ and ethyl are on opposite sides of the double bond, so the configuration is $(E)$. Tetrasubstituted alkenes do show $E/Z$ isomerism whenever each carbon carries two different groups; the positions of the two methyls are irrelevant to the CIP label.