The shape is about right for the variance of the M/M/1 queue (what is an M/M/1/∞ queue??), but it's a little disingenuous since variance is a higher-order moment. Standard deviation is probably more sensible to think about here, since that brings it back down to the same scale as the queue length.
> Thus the average number of customers in the system is ρ/(1 − ρ) and the variance of number of customers in the system is ρ/(1 − ρ)². This result holds for any work conserving service regime, such as processor sharing.
In the cited example, going from 75% utilization -> 95% utilization drags your standard deviation from a modest 3.5 items in the queue all the way up to 19.5 items. And, in both cases, that's not far off from your average queue length, either.
Infinite queue length. M/M/1/k would be a system where once k items are enqueued, some work shedding policy takes over (either rejecting new work or often preferably dropping at the head of the queue.)
> Thus the average number of customers in the system is ρ/(1 − ρ) and the variance of number of customers in the system is ρ/(1 − ρ)². This result holds for any work conserving service regime, such as processor sharing.
https://en.wikipedia.org/wiki/M/M/1_queue#Average_number_of_...
In the cited example, going from 75% utilization -> 95% utilization drags your standard deviation from a modest 3.5 items in the queue all the way up to 19.5 items. And, in both cases, that's not far off from your average queue length, either.