Because each isotope contributes equally to the total mass regardless of abundance, the resulting average atomic mass is simply the median value, which does not align with $$80.2$$ amu.

Since the isotope with mass $$80$$ constitutes the majority of the sample, its high abundance drives the weighted average atomic mass to be close to $$80.2$$ amu.

The weighted average must be lower than $$80.2$$ amu due to the contribution of the lighter isotope, Se-78, despite its lower abundance.

Because the atomic masses span from $$78$$ to $$82$$, the average atomic mass must be exactly the arithmetic mean of these values, namely $$80$$ amu.