Average passanger waiting time - probability density function - normal distribution

What your professor mentioned is likely "renewal theory", which deals with such things. Your waiting time is here the "residual time" in this theory. See https://en.wikipedia.org/wiki/Renewal_theory#Renewal_function and https://en.wikipedia.org/wiki/Residual_time for explanations and formulas. The scope of this answer is too limited to derive the formulas used. But all formulas are in these two wikipedia pages.

What I can do is write down the general expressions, which are intractable integrals. I can then take the "large time" to get specific numbers. These numbers are somewhat different from intuition and the case for fixed arrival times.

One final remark: Assuming that the intermediate times are normal distributed means that negative intermediate times are also possible. This is clearly impossible, so we should "restrict and rescale" the normal distribution to interval $(0,\infty)$. But when doing integrals its much better to ignore this problem of negative times and take the domain to be $(-\infty, \infty)$.

1. Suppose that the passenger arrives at time $t$, the probability density for the passenger to wait time $T$ is given by:
$$p(w=T\mid a = t) = \mathcal N(t+T;\mu,\sigma) + \int_0^t \mathcal N(T+\tau; \mu,\sigma) \cdot m'(t-\tau)d\tau$$
(compare: https://en.wikipedia.org/wiki/Residual_time#Probability_distribution_of_the_residual_time ) whence the probability density for the waiting time independent of the arrival is:
$$p(w=T) =\frac1{end\ of \ day} \int_0^{end\ of\ day }p(w=T\mid a=t) dt$$
here $\mathcal N(x;\mu,\sigma)$ is the probability density of the Gaussian with parameterse $\mu, \sigma$. The function $m'$ is something more complicated, it is the derivative of $m(t) = \Bbb E[N(t)]$ where $N(t)$ is the number of busses that have appeared at time $t$. This becomes an intractable expression, it can only be simplified in the limit of large $t$.

For large $t$ you have that $m'(t)\approx \frac1\mu$, see eg the renewal theorem or "key renewal theorem" on the wikipedia page. This fact is however very intuitive. If you put it in the first equation you get that:
$$p(w=T\mid a=t_{large}) \approx 0 + \int_0^{t_{large}} \mathcal N(T+\tau;\mu,\sigma)\frac1\mu d\tau\approx \frac1\mu\int_{T}^\infty \mathcal N(\tau;\mu,\sigma)d\tau$$

2. The expected waiting time is given by
$$\int_0^{\infty} T p(w=T)\,dT $$
we approximate $p(w=T)$ by $p(w=T\mid a = t_{large})$, since with an intermediate arrival time of $4$ minutes we will have very many arrivals throughout the day, so most of the time a large number of busses have already arrived and the integral giving $p(w=T)$ basically contains only terms where $t$ is large. Then we can approximate this by:
$$\Bbb E[w] \approx \int_0^\infty T \frac1\mu (1-\int_{-\infty}^T \mathcal N(\tau;\mu,\sigma) d\tau) \ dT$$
this integral can be evaluated, use partial integration on the term of the form:
$$\int_0^{V}x (\int_{-\infty}^x f(y) dy)dx = [\frac12 x^2 \int_{-\infty}^x f(y)dy]_{x=0}^{x=V} - \int_0^V \frac12 x^2 f(x) dx$$
then:
$$\frac1\mu\int_0^V(T- T\int_{-\infty}^T \mathcal N(\tau) d\tau)=\frac1\mu \left[\frac12 T^2 \underbrace{(1-\int_{-\infty}^T \mathcal N(\tau))}_{\to0\ as\ T\to\infty}\right]_{T=0}^{T=V}+ \frac1{2\mu}\int_{-\infty}^V \tau^2 \mathcal N(\tau)d\tau$$
for $V\to\infty$ we recover the integral for $\Bbb E[w]$. It then becomes $\frac1{2\mu_1}\mu_2$ where $\mu_1=\mu$ and $\mu_2$ is the second moment of the Gaussian, ie the expectation of $x^2$, which is $(\mu^2+\sigma^2)$. The expected waiting time is then
$$\Bbb E[w]\approx \frac{\mu^2+\sigma^2}{2\mu}$$
which is larger than the expected waiting time in the fixed departure case.

3. The standard deviation is $\sqrt{\Bbb E[w^2]-\Bbb E[w]^2}$. One can use similar tricks as above to get this value:
$$\sqrt{\frac{\mu^2+3\sigma}{3}-\frac{(\mu^2+\sigma^2)^2}{4\mu^2}}$$

4. As above for large times we use our approximation
$$p(w=T\mid a = t_{large})\approx \frac1\mu(1- \int_{-\infty}^T \mathcal N(\tau) d\tau)$$
the probability to wait a time larger than $5$ minutes is then
$$\int_{5}^\infty p(w=T\mid a=t_{large}) dT $$
we can evaluate this numerically or with a computer software if we want. The software tells us that we get
$$\frac{1}{2} \text{erfc}\left(-\frac{1}{25} \left(18 \sqrt{2}\right)\right)+\frac{25}{36 e^{648/625} \sqrt{2 \pi }}-1$$
which is approximately $0.0233$.