From Flat Random Numbers to Arrival Times

A help-desk simulator needs realistic times at which customers arrive. The gap $X$, in minutes, between one arrival and the next is a continuous random variable with CDF

where $\mu$ (mean_gap) is the average gap in minutes.

Two tools you will need:

• $e \approx 2.71828$ is a fixed number, like $\pi$. $e^{a}$ means $e$ raised to the power $a$; in Python it is math.exp(a).
• The natural logarithm $\ln$ undoes it: for any number $a$, $\ln(e^{a}) = a$. So if $e^{a} = c$ for a positive number $c$, then $a = \ln(c)$. In Python, $\ln(c)$ is math.log(c).

The simulator's random-number generator can only produce uniform draws: numbers $u$ between 0 and 1, with every value equally likely. To turn a draw into a gap, it reads the CDF in reverse. The gap for draw $u$ is the value $x$ at which the CDF reaches $u$:

This works because $F$ climbs from 0 towards 1 and never goes back down, so each $u$ picks out exactly one $x$. Where $F$ climbs steeply (gaps that are likely), a wide band of $u$ values all land on a short stretch of $x$, so many draws give gaps there. Where $F$ climbs slowly (gaps that are rare), a thin band of $u$ values is spread over a long stretch of $x$, so few draws give gaps there. The gaps produced this way have exactly the CDF $F$.

The first customer arrives one gap after time 0, and each later customer arrives one gap after the customer before.

Task: write arrival_times(uniforms, mean_gap) that turns each draw, in order, into a gap and returns the list of arrival times (minutes since time 0), each rounded to 4 decimal places.

• Every draw satisfies $0 \le u < 1$, and mean_gap is positive.
• Use the draws exactly as given: the $k$-th draw produces the $k$-th gap.
• Round only the arrival times you return, not the gaps along the way.
• An empty list of draws means no customers, so return [].