Let $R$ be a commutative ring with unity such that every non-zero module over $R$ has an associated prime. Then is it true that $R$ is Noetherian ? If not, then can we say something about any possible structure of $R$ at least ?

COMMENTS: Since for every proper ideal $I$ of $R$, and non-zero $x+I \in R/I$, $\mathrm{ann}_R(x+I)=(I:x)$, so if every non-zero module over $R$ has an associated prime, then for every proper ideal $I$ of $R$, $\exists x \notin I$ such that $(I:x)$ is a prime ideal, and conversely, if this holds, then for every proper ideal $I$ of $R$, $\mathrm{Ass}(R/I)$ is non-empty, and then since for every non-zero $R$-module $M$, we have $\mathrm{Ass}(R/\mathrm{ann}(m)) =\mathrm{Ass}(Rm)\subseteq \mathrm{Ass}(M)$, for every non-zero $m$ in $M$, so every non-zero $R$-module would have an associated prime. So we have proved :

For a commutative ring (with unity) $R$, TFAE :

1) for every proper ideal $I$ of $R$, $\exists x \notin I$ such that $(I:x)$ is a prime ideal.

2) $R/I$ has an associated prime for every proper ideal $I$ of $R$.

3) Every non-zero $R$-module has an associated prime.

But condition (1) still seems very complicated. Noetherian rings of course satisfy condition (1), but I don't know what other type of rings satisfy that condition.