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The Kaplan–Meier estimator is $\hat{S}(t)={\prod }_{t_j\le t}(1-d_j/r_j)$, with event count $d_j$ and risk set size $r_j$. Its median is the first event time with estimated survival at most $1/2$. Since $\hat{S}(2)=0.5749$ and $\hat{S}(3)=0.4286$, the estimated median completion time is $3$ minutes.

The original last event exhausts the final risk set, so the fitted survival function drops to zero at $18$ and its conventional full area is finite. Integrate the right-continuous step function: its height is one before time one, and after each event the height is the newly computed survival value. There are three-minute gaps from $12$ to $15$ and from $15$ to $18$. ThusUsing the exact risk-set products rather than rounded printed survival values gives $4.187856$. This is not the sample mean of the observed event-or-censoring times; right censoring changes the estimated survival weights.

If the final observation at $18$ is instead censored, every earlier Kaplan–Meier estimator value is unchanged and no event factor is inserted at $18$. The median remains $3$ minutes, but the curve stays at $\hat{S}(15)\simeq 0.0107$ at the end of follow-up. By terminal censoring and survival-mean identifiability, a full unrestricted mean is no longer determined nonparametrically by these observations. Extending the final positive step forever would give an infinite integral, but that plotting convention does not establish an infinite population mean. Extra assumptions about the unobserved tail would be needed for a finite full-mean estimate.

The restricted mean survival time through $18$ is still $4.187856$ minutes in either version: the two curves differ only at or after the endpoint, which does not affect the area up to it. All these interpretations require independent censoring. Giving up may be related to how long a person would have taken to finish, so that assumption needs scientific justification in this study.

Let $a_1<a_2<\cdots$ be the distinct observed event times, $r_j$ the risk set size immediately before $a_j$, and $d_j$ the events there. Under independent censoring, the Kaplan–Meier estimator isIt starts at one, jumps down at events, and is unchanged by a pure censoring time; right censoring removes people from subsequent risk sets. If events and censorings coincide, the convention here processes events before censoring, so those censored at the recorded time are included just before the event. The estimated median is the first time the curve reaches or falls below $1/2$. If it never does during observation, the median is not estimable from the observed curve without tail assumptions.

With one remaining individual censored at $t_1$, no event factor is introduced. HenceThis is the terminal observed step value. With an empty risk set after $t_1$, continued horizontal plotting is a convention and supplies no information about the true later survivor function. The tail rests on one person and is imprecise; this is terminal censoring and survival-mean identifiability. If ${\varphi }_0>0$, the unrestricted mean cannot be obtained nonparametrically from this censored tail: extending the last positive step to infinity would give an infinite area, which does not establish an infinite population mean. A restricted mean survival time up to a suitably supported finite cutoff remains estimable.

For a nonnegative survival time $T$, the survival function iswhere $F$ is the cumulative distribution function. Thus $S(t)$ is the probability that the chosen occurrence is still in the future at time $t$. We consistently use the strict inequality $T>t$: a probability mass at $t$ has already left the survival function at $t$ itself. With this convention $S$ is a right-continuous function and a nonincreasing function taking values in $[0,1]$. If $\mathbb{P}(T=0)=0$, then $S(0)=1$; if $T$ is finite with probability one, then $S(t)\to 0$ as $t\to \infty$.

For $S(a)>0$ and $b>a$, conditional probability givesFor example, $S(2)=0.8$ and $S(5)=0.6$ mean that $0.6/0.8=0.75$ of those still event-free at year two remain event-free through year five. Their intervening event probability is $0.25$, whereas the original population's probability of an occurrence in $(2,5]$ is $0.8-0.6=0.2$. Conditioning changes the denominator.

The area under the survival function up to a horizon gives the restricted mean survival time. A positive terminal plateau in a plotted Kaplan–Meier estimator does not establish that the occurrence will never happen: terminal censoring and survival-mean identifiability explains why the unobserved tail remains undetermined.