Suppose customers arrive at a system as a Poisson process with rate $\lambda$, given a specific time interval $[0,t]$, what is the expected wait times for those customers who arrive in this interval?
To clarify, let the number of customers arriving in $[0,t]$ be $N$, and their arrival times are $T_1, T_2, \cdots, T_N$ respectively where $0\le T_1, T_2, \cdots, T_N\le t< T_{N+1}$. For each of them, the wait time within the interval $[0,t]$ is $t-T_i$. The average wait time is, $$\bar{W}=\frac{\sum^N_1{t-T_i}}{N}$$
Since $N$ is an r.v., so is $\bar{W}$. The expectation of $\bar{W}$ is $$E[\bar{W}]=\sum^\infty_1 \bar{W}(N,t)\cdot p(N)$$
However $T_i$'s are r.v.'s too, it turns out to be kind of complicated for me. Could someone help me out? thanks.
EDIT Here is an updated version which matches the result from symmetry in the comment on the OP:
Let $N_t \sim Pois(\lambda t)$ and condition on the event of $n$ arrivals in time $(0,t]$, i.e. $N_t=n$. Then it is relatively well known that the arrival times, $T_i$, are distributed like the order statistics of $n$ uniformly distributed RVs on $(0,t]$. That is $T_i \sim U_{(i)}$ where $U_{(i)}$ is the $i$-th order statistics of $n$ IID $\mathcal{U}(0,t)$ RVs, $U_1,\dotsc, U_n$ (e.g. $U_{(1)}:=\min\{U_1,\dotsc, U_n\}$, $U_{(n)}:=\max\{U_1,\dotsc, U_n\}$ and $U_{(i)}$ is the $i$-th smallest number of $\{U_1,\dotsc, U_n\}$).
Thus, we have $\mathbb{E}(T_i | N_t=n)=\mathbb{E}(U_{(i)})$, so it remains to compute these means in order to compute the expectation of $\bar{Y}_n=t-\frac{1}{n}(T_1+\dotsc +T_n).$ I will write out the full computation of $\mathbb{E}(U_{(1)})$ since by your comment these order statistics are unfamiliar but I will leave the rest for you to go through yourself.
So, simply by the definition of minimum we have the following set equality, $$\{U_{(1)} \geq u\}=\{U_1 \geq u, \dotsc, U_n \geq u\},$$ hence, by the fact that all the $U_i$ are IID, $$\mathbb{P}(U_{(1)}\geq u)=\mathbb{P}(U_1 \geq u)^n=(1-u/t)^n.$$
Now we use the wonderful fact that for every non-negative random variable $X$, we may compute the expected value in an alternate fashion, as $\mathbb{E}(X)=\int \mathbb{P}(X\geq x) \mathrm{d}x,$ (omitting limits of integration for brevity), so that in this case, $$\mathbb{E}(U_{(1)})=\int_0^t (1-u/t)^n \mathrm{d}u=\frac{t}{n+1}.$$
Using an analogous argument we can compute $\mathbb{E}(U_{(n)})=\frac{nt}{n+1}$. Then, further, I claim $\mathbb{E}(U_{(i)})=\frac{it}{n+1}$ for all $i=1,2,\dotsc,n$. Thus, we see that (after some algebra) $$\mathbb{E}(T_1+\dotsc+T_n |N_t=n)=\mathbb{E}(U_{(1)}+\dotsc+U_{(n)})$$ $$=\frac{nt}{2},$$ from which it immediately follows that $\mathbb{E}(\bar{Y}_n | N_t=n)=t-\frac{t}{2}=\frac{t}{2}$. Now since this is just a constant independent of $N_t=n$, we finally get by the law of total expectation, $$\mathbb{E}(\bar{Y}_{N_t})=\mathbb{E}\mathbb{E}(\bar{Y}_{N_t} | N_t)=\frac{t}2,$$ just as user @Henry claimed by symmetry.
Original answer that is incorrect
So, conditional on $N_t=n$ customers, $Y_1+\dotsc+Y_n=nt-(T_1+\dotsc+T_n)$, hence the sample mean is $\bar{Y}_n=t-\frac{1}{n}(T_1+\dotsc+T_n)$, which has conditional expectation $\mathbb{E}(\bar{Y}_n | N_t=n)=t-\frac{n+1}{2\lambda}.$ Here, we used the fact that $$\mathbb{E}(T_1+\dotsc+T_n)=\frac{1}{\lambda}+\frac{2}{\lambda}+\dotsc+\frac{n}{\lambda}=\frac{1}{\lambda}(1+2+\dotsc +n)=\frac{n(n+1)}{2\lambda}.$$ That is to say, conditional on knowing $N_t$, $\mathbb{E}(\bar{Y}_{N_t} | N_t)=t-\frac{N_t+1}{2\lambda},$ (conditional expectations are, in fact, RVs!).
Hence, upon taking the expectation with respect to $N_t$, we get, by the law of total expectation, $$\mathbb{E}(\bar{Y}_{N_t})=\mathbb{E}\mathbb{E}(\bar{Y}_{N_t} | N_t)=t-\frac{1}{2\lambda}\mathbb{E}(N_t+1)=\frac12 \left(t-\frac{1}{\lambda}\right),$$ just using $\mathbb{E}(N_t)=\lambda t$ and linearity.