Awesome averages.

Since all digits are used exactly once in each number, the possible arrangements of the digits will result with all the numbers in each digit place the same amount of times. Therefore, the averages of each digit place will be the same. These averages can be calculated by adding up the digits and dividing the sum by 5. The resulting sum is 4.8. When entered into each digit place, 48000 + 4800 + 480 +48 +4.8 = 53332.8 is found to be the answer.

Let us look at all such numbers starting in $1$ : there are $4!$ such numbers, and so the sum of all the first (leftmost) digits is $4! \times 1$ .

By a similar argument, the sum of all the $n$ th digits of all such numbers with $k \in \lbrace 1,3,5,7,8 \rbrace$ in the $n$ th place is $4! \times 10^{n-1}k$ .

Altogether, there are $5!$ such numbers.

Hence, the average is

$\displaystyle \frac{1}{5!} \sum_{n=1}^{5} 4! \times 10^{n-1}k = \frac{11111+33333+55555+77777+88888}{5}$ $= \boxed{53332.8}$