You do not give much information about these waiting times, so this answer has to be regarded as highly speculative.
Often waiting times are well-modeled as exponential. Your small sample $0.2, 1.0, 0.3, 0.0, 0.1$ has high outlier $1.0$ which may suggest such a right skewed distribution. A sample of size five is hardly enough for reliable speculation about your distribution of waiting times.
If $\bar X$ is the mean of an exponential population of size $n,$ one has $\frac{\bar X}{\mu} \sim \mathsf{Gamma}(\mathrm{shape}=n,\mathrm{rate}=n).$ This relationship can be 'pivoted' to give a 95% CI for $\mu$ of the form $\left(\frac{\bar X}{U}, \frac{\bar X}{L}\right),$ where $L$ and $U$ cut area 0.025 from the lower and upper tails of $\mathsf{Gamma}(n,n),$ respectively.
So if you have $n = 20$ exponential observations with $\bar X = 0.35,$ then a 95% CI for $\mu$ is $(0.24, 0.57).$ [Computations in R.]
0.35/qgamma(c(.975,.025),20,20)
[1] 0.2359218 0.5729946
Taking the worst-case scenario that the mean is as large as $\mu = 0.57,$ one might speculate that 99% of the time the waiting time would be less than $2.624.$
qexp(.99, 1/0.57) # mean 0.57 implies rate 1/0.57
[1] 2.624947
Of course, this is based on only 20 observations, the assumption that future waiting times will imitate past ones, and the assumption that waiting times are exponential.
By contrast, if you have a substantial amount of data on relevant past waiting times, then you have more information and fewer assumptions. You might use the 99th percentile of that data for a 'pessimistic' prediction of the next waiting time, as shown below.
set.seed(2021)
y = rexp(1000, 1/0.35) # fictitious exponential data
for illustration
summary(y)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.0002822 0.0990812 0.2418578 0.3520501 0.4953001 2.3407786
quantile(y, .99)
99%
1.535914
Based on my fictitious data, only ten times in 1000 such situations was the waiting time more than $1.536$ and the waiting time was never more than $2.34.$$2.35.$
sum(y > 1.536)
[1] 10