3. A set of mutually unequal data has a median of 80, the average of the numbers less than the median is 70, and the average of the numbers greater than the median is 96. Let the average of this set of data be $\bar{x}$, then $($.
(A) $\bar{x}=82$
(B) $\bar{x}=83$
(C) $80 \leqslant \bar{x} \leqslant 82$
(D) $82 \leqslant \bar{x} \leqslant 83$
3.D.

If this set of data has $2 n$ numbers, then
$$
\bar{x}=\frac{70 n+96 n}{2 n}=83 ;
$$

If this set of data has $2 n+1$ numbers, then
$$
\bar{x}=\frac{70 n+80+96 n}{2 n+1}=83-\frac{3}{2 n+1} \text {. }
$$

Since $n \geqslant 1$, therefore, $82 \leqslant \bar{x} \leqslant 83$.