Let $k$ be the fourth largest number. Then it must satisfy $3 < k < 90$. Meanwhile, for $k$ to be the fourth largest number, there must be three numbers chosen from $1$ to $(k-1)$, inclusive, and one number chosen from $(k+1)$ to $90$, inclusive. Therefore, there are totally $(90-k)\binom{k-1}{3}$ different ways this scenario can occur. It follows that the expected value is $$E=\frac{1}{\binom{90}{5}}\underbrace{\sum_{k=1}^{89}k(90-k)\binom{k-1}{3}}_{A}$$

where $\binom{90}{5}$ is the number of different ways to choose $5$ distinct number from $90$ candidates.

By the basic combinatorial identity, we have $$k\binom{k-1}{3}=4\binom{k}{4}$$

Therefore $$\begin{align*} A =\ &\sum_{k=1}^{89}k(90-k)\binom{k-1}{3} \\ =\ & 4\sum_{k=1}^{89}(90-k)\binom{k}{4}\\ =\ & 4\sum_{k=1}^{89}(91-(k+1))\binom{k}{4} \\=\ &4\cdot 91\underbrace{\sum_{k=1}^{89}\binom{k}{4}}_{B}-4\underbrace{\sum_{k=1}^{89}(k+1)\binom{k}{4}}_{C}\end{align*}$$

Part $B$ above can be evaluated using the hockey stick identity $$B=\sum_{k=1}^{89}\binom{k}{4} = \sum_{k=4}^{89}\binom{k}{4}=\binom{90}{5}$$

Part $C$ can be transformed again by identity $(k+1)\binom{k}{4}= 5\binom{k+1}{5}$ before applying the hockey stick identity $$C=\sum_{k=1}^{89}(k+1)\binom{k}{4}=5\sum_{k=1}^{89}\binom{k+1}{5}=5\cdot\binom{91}{6}$$

Therefore, $$A=4\cdot 91\cdot\binom{90}{5} - 4\cdot 5\cdot\binom{91}{6}=\frac{182}{3}\binom{90}{5}$$

It follows that the answer is $E=\boxed{\frac{182}{3}}$