Find the average value!

Let the average value of $|a_i-a_j|$ be $\overline{n}$ . Then, the average value of $|a_1-a_2| + |a_3-a_4| + |a_5-a_6| + |a_7-a_8| + |a_9-a_{10}|$ is $5\overline{n}$ . Let $n_k$ be the number of $|a_i-a_j| = k$ .

For $k=1$ , we have $|\pm 1\mp 2|$ , $|\pm 2\mp 3|$ , $|\pm 3\mp 4|$ , $|\pm 4\mp 5|$ , $|\pm 5\mp 6|$ , $|\pm 6\mp 7|$ , $|\pm 7\mp 8|$ , $|\pm 8\mp 9|$ and $|\pm 9\mp 10|$ . Therefore, $n_1 = 18$ . Similarly, $n_2 = 16$ , $n_3 = 14$ , $\Rightarrow n_k = 2(10-k)$ .

We know that $\overline{n}$ is the expected value of $n_k$ and it is given by:

$\begin{aligned} \overline{n} & = \dfrac{\displaystyle \sum_{k=1}^9 k n_k} {\displaystyle \sum _{k=1} ^9 n_k} = \dfrac{\displaystyle \sum_{k=1}^9 2k(10-k)} {\displaystyle \sum _{k=1} ^9 2(10-k)} = \dfrac{\displaystyle \sum_{k=1}^9 k(10-k)} {\displaystyle \sum _{k=1} ^9 k} = 10 - \dfrac{\displaystyle \sum_{k=1}^9 k^2} {\displaystyle \sum _{k=1} ^9 k} \\ & = 10 - \frac{9(10)(19)}{6} \times \frac{2}{9(10)} = 10 - \frac{19}{3} = \frac{11}{3} \end{aligned}$

Therefore, the required answer $5\overline{n} = 5 \times \dfrac{11}{3} = \boxed{18.33}$ to two decimal places.