At month 12, patients C and D are censored at durations 6 and 1, respectively. When B dies at duration 7, only A and B remain at risk, so
$$
\widehat F_{12}(12)=1-\frac12=\frac12.
$$

At month 36, all six patients have at least twelve months of potential follow-up unless they die earlier. The only deaths by duration twelve are B and E, tied at duration seven, so
$$
\widehat F_{36}(12)=1-\frac26=\frac23.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 207 / 5 / v / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-207/5/v/solution