**EDIT:** After some discussion, the OP made it clear that they were actually asking about a more fundamental issue: given a time-dependent probability density $p(x,t)$, and given that we are observing a fixed spatial interval, when do we expect to _first_ observe the event? 

(Only the first observation is important, because the detection of a particle is an interaction that changes its wavefunction, and so we stop wondering _when_ we'll detect the particle once we actually _do_ detect the particle). 

Let's ask a simpler question first, that might guide our intuition. Let's roll a die. The outcomes are 1 to 6, all equally probable, and each roll of the die is a discrete time interval (let's say we roll once per second). Let's ask the question: how long will it take, on average, for us to roll a 4?

The probability of rolling a 4 on the first roll is $1/6$. The probability of rolling your first 4 on the second roll and _not_ on the first roll is $1/6\times(1-1/6)$. Likewise, the probability of rolling a 4 on the third roll but _not_ on the first or second is $1/6\times(1-1/6)^2$. And the probability of rolling a 4 on the $n$th roll but _not_ on any previous roll is $1/6\times (1-1/6)^{n-1}$. So, from our original probability distribution of outcomes per time interval, we can assemble a probability distribution of the amount of time it will take us to see a 4:

$$P(t_n)=1/6\times(1-1/6)^{n-1}$$

where $t_n$ is the $n$th time interval. The mean value of $t_n$, the expected time interval in which we'll see our first 4, is:

$$\bar{t}=\sum_{n=1}^\infty nP(t_n)=\sum_{n=1}^\infty n\times 1/6\times (1-1/6)^{n-1}=6$$

So we should expect it to take roughly 6 seconds to see our first 4.

With a few tweaks, we can apply that logic to our current situation. Suppose we're observing over the spatial interval $a<x<b$. First, we need to calculate the probability of observing our outcome as a function of time:

$$P(t)=\int_{a}^b p(x,t) dx$$

Now, we discretize our continuous time parameter. Our detector interacts with the environment, but those interactions are not instantaneous: every interaction that would allow a detection has some associated timescale $\Delta t$ (for example, detectors based on ionization would have a timescale associated with the amount of time an incoming particle takes to ionize an atom). So we can model our detector as a device that periodically "checks" to see whether it interacted with a particle. So now we have a set of discrete time intervals, $t=0, \Delta t,2\Delta t,...$ during which the metaphorical dice are rolled. 

But this time, each time these metaphorical dice are rolled, the probability is different. And it's clear that we can't actually use the probability at a particular instant, either, because that would imply that we know what the "phase" of the detector's interactions are, which we don't. So instead, we _average_ the probability over one interaction timescale. Let $P_n$ be the probability that a detector detects a particle in the interaction timescale interval $(n\Delta t, (n+1)\Delta t)$:

$$P_n=\frac{1}{\Delta t}\int_{n\Delta t}^{(n+1)\Delta t} P(t)dt$$

So we can now play the same game as before: the probability that we detect a particle on the very first interaction timescale is $P_0$. The probability that we detect a particle on the second interaction timescale but _not_ the first one is $P_1(1-P_0)$. The probability that we detect a particle on the third interaction timescale but not the second or first is $P_2(1-P_1)(1-P_0)$. And so on, generating our formula for the probability of seeing our particle on the $n$th interaction timescale:

$$P(\text{detection after }n\text{ interaction timescales})=P_n(1-P_{n-1})(1-P_{n-2})...(1-P_1)(1-P_0)$$

Now that we have our distribution for arbitrary $n$, this means that the expected number of interaction timescales that we'll have to wait to detect the particle is:

$$\bar{n}=\sum_{n=0}^\infty nP_n(1-P_{n-1})(1-P_{n-2})...(1-P_0)$$

Once we have numerically calculated $\bar{n}$, then we can easily get the expected wait time before detecting a particle:

$$\bar{t}=\bar{n}\Delta t$$

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With that out of the way, let's calculate the actual probability density function.

Let's suppose that you prepare your Gaussian wavepacket in a _minimum-uncertainty_ configuration. What I mean by that is described below.

The Heisenberg uncertainty principle states:

$$\sigma_x\sigma_p\geq\frac{\hbar}{2}$$

It turns out that the situation where the product $\sigma_x\sigma_p$ is _minimized_ is actually a Gaussian wavefunction (proofs of this can be found elsewhere on the internet), so for that particular Gaussian wavefunction, we have:

$$\sigma_x\sigma_p=\frac{\hbar}{2}$$

The momentum probability distribution is also Gaussian, with some mean $\bar{p}$ and a standard deviation $\sigma_p=\frac{\hbar}{2\sigma_x}$.

So if we start with our Gaussian momentum wavefunction $\psi(k)=e^{-\alpha(k-k_0)^2}$, where $\alpha=\frac{\hbar^2}{2\sigma_p^2}=\sigma_x^2$, we can follow [this procedure][1] to find the position wavefunction as a function of time (and then normalize said wavefunction, because the authors of that source apparently didn't bother to do so):

$$\psi(x,t)=\frac{1}{(2\pi)^{1/4}\sqrt{\alpha+i\beta t}}e^{i(k_0x-\omega_0 t)}e^{\frac{-(x-v_g t)^2}{4(\alpha+i\beta t)}}$$

where $v_g=\frac{d\omega}{dk}$ evaluated at $k_0=\frac{\bar{p}}{\hbar}$, and $\beta=\frac{1}{2}\frac{d^2\omega}{dk^2}$, also evaluated at $k_0$. 

As you can see, in order to proceed, we need a relation between $\omega$ and $k$. This is called the _dispersion relation_, and for a relativistic electron, the dispersion relation is:

$$\omega=c\sqrt{k^2+(m_ec/\hbar)^2}$$

This means that:

$$\omega_0=c\sqrt{k^2+(m_ec/\hbar)^2}$$

$$v_g=\frac{ck_0}{\sqrt{k_0^2+(m_ec/\hbar)^2}}$$

$$\beta=\frac{c}{2\sqrt{k_0^2+(m_ec/\hbar)^2}}-\frac{ck_0^2}{2(k_0^2+(m_ec/\hbar)^2)^{3/2}}$$

Then, figuring out the probability that the electron will be at the screen position $x_s$ as a function of time is as simple as evaluating $|\psi(x_s,t)|^2$:

$$|\psi(x_s,t)|^2=\frac{1}{\sqrt{2\pi(\alpha^2+\beta^2t^2)}}\exp\left(\frac{-\alpha(x_s-v_gt)^2}{2(\alpha^2+\beta^2t^2)}\right)$$

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Obviously, this general solution doesn't tell us mere mortals very much in terms of intuition, so there are two special cases that are helpful to develop some understanding of the situation:

**The ultra-relativistic limit**

In the case where $k\gg m_ec/\hbar$, the dispersion relation reduces to:

$$\omega=ck$$

which means:

$$\omega_0=ck_0$$

$$v_g=c$$

$$\beta=0$$

Plugging these into the general solution, we find that:

$$|\psi(x_s,t)|^2=\frac{\pi}{\sigma_x^2}\exp\left(-\frac{(x_s-ct)^2}{2\sigma_x^2}\right)$$

As you can see, the wavefunction simply travels to the right at velocity $c$ over time, with a constant width $\sigma_x$ as a function of time. So the uncertainty in detection time depends only on the uncertainty in initial position of the electron.

**The non-relativistic limit**

In the limit where $k\ll m_ec/\hbar$, the dispersion relation reduces to:

$$\omega\approx \frac{m_ec^2}{\hbar}+\frac{\hbar k^2}{2m_e}$$

which means that:

$$\hbar\omega_0=m_ec^2+\frac{p^2}{2m_e}$$

$$v_g=\frac{\hbar k_0}{m}=\frac{\bar{p}}{m}$$

$$\beta=\frac{\hbar}{2m}$$

Plugging these into the original formula, we find that the center of the wavepacket travels with a velocity $v_g$, as you would expect, and that the wavepacket also spreads out quite a bit over time: the width of the wavepacket is $\sqrt{\alpha^2+\left(\frac{\hbar t}{2m}\right)^2}$. So the uncertainty in the detection time depends both on the initial uncertainty in position _and_ on the distance from the mean initial position to the screen. Generally, the further away the screen is, the more uncertain the detection time will be.

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With these two extremes, we can now interpolate between them to say something about what happens to a relativistic (but not ultra-relativistic) electron: increasing the distance to the screen still increases the uncertainty in detection time, but not by as much as in the non-relativistic case (which makes sense - at relativistic speeds, changing your momentum doesn't actually change your velocity very much). 

Incidentally, this is why time-of-flight detectors in particle physics experiments only work well at lower energies: determining momentum by measuring velocity gets more and more difficult as energy increases.

  [1]: https://quantummechanics.ucsd.edu/ph130a/130_notes/node83.html