Let $R$ be a Noetherian ring and $M$ a finite $R$-module. Then the minimal primes of $M$, i.e. the minimal associated primes of $M$, coincide with the minimal primes of the ring $R/\operatorname{ann}(M)$.
However, it is not true in general that $\operatorname{Ass}(M) = \operatorname{Ass}(R/\operatorname{ann}(M))$, right?
Take $R$ a noetherian integral domain and $M=R\oplus R/P$ with $P$ a non-zero prime.
Remark. Let me mention that we have $\operatorname{Ass}(R/\operatorname{Ann}(M))\subseteq\operatorname{Ass}(M)$. This follows easily: if $M=Rx_1+\cdots+Rx_n$, then we have $0\to R/\operatorname{Ann}(M)\to\oplus_{i=1}^nRx_i$, so $$\operatorname{Ass}(R/\operatorname{Ann}(M))\subseteq\operatorname{Ass}(\oplus_{i=1}^nRx_i)=\cup_{i=1}^n\operatorname{Ass}(Rx_i)\subseteq\operatorname{Ass}(M).$$