Solution. \(\hat S(12) = (1 - \tfrac 1{10})(1 - \tfrac 19)(1 - \tfrac 17)(1 - \tfrac 16) = 0.9 \times 0.8889 \times 0.8571 \times 0.8333 = 0.5714\). The Greenwood sum is \begin {align*} &\frac {1}{10 \times 9} + \frac {1}{9 \times 8} + \frac {1}{7 \times 6} + \frac {1}{6 \times 5} \\ &\qquad = 0.0111 + 0.0139 + 0.0238 + 0.0333 = 0.0821 , \end {align*}

so \(\text {SE} = 0.5714\sqrt {0.0821} = 0.164\).

Solution. Group B: at risk \(10, 9, 8\) at \(2, 3, 4\) months (one event each), then the patient censored at \(5\) leaves, so \(6\) at risk at \(6\) months. \(\hat S(6) = 0.9 \times \tfrac 89 \times \tfrac 78 \times \tfrac 56 = 0.583\). Then \(\tfrac 45\) at \(8\) and \(\tfrac 34\) at \(10\): \(\hat S(10) = 0.583 \times 0.8 \times 0.75 = 0.350\). After the censoring at \(11\), two remain; events at \(14\) and \(16\) give \(\hat S(16) = 0.350 \times \tfrac 12 \times 0 = 0\). The median is the first time \(\hat S \leq 0.5\): \(8\) months for B (\(\hat S = 0.467\)) and \(18\) months for A (\(\hat S = 0.429\)).

Solution. With \(n_i' = l_i\): \(q = 0.0800, 0.0795, 0.0789, 0.0625, 0.0577\), giving \(\hat S_5 = 0.689\). This treats the withdrawn as at risk for the whole interval, so it overestimates survival. With \(n_i' = l_i - w_i\): \(q = 0.0833, 0.0843, 0.0857, 0.0714, 0.0698\), giving \(\hat S_5 = 0.663\), which treats them as at risk for none of it and underestimates survival. The actuarial value \(0.677\) lies between, assuming withdrawal half-way through on average.

Solution. Entry is at age \(45.5\) and death at exactly \(57\). Person-time: \(4.5\) years at \(45\)–\(49\) (to age \(50\), on 15 March 1990), \(5\) years at \(50\)–\(54\), and \(2\) years at \(55\)–\(59\); total \(11.5\). The death, at the start of the person’s \(58\)th year, is counted in the \(55\)–\(59\) band.

Solution. \(l_{60} = e^{-0.12} = 0.887\) and \(l_{80} = 0.887\,e^{-0.4} = 0.594\). Under a constant hazard \(m\), the expected time lived in an interval of length \(n\) by a person alive at its start is \((1 - e^{-nm})/m\), and above \(80\) it is \(1/m\). So \begin {align*} e_0 &= \frac {1 - e^{-0.12}}{0.002} + 0.887 \times \frac {1 - e^{-0.4}}{0.02} + 0.594 \times \frac {1}{0.15} \\ &= 56.5 + 14.6 + 4.0 = 75.1 \end {align*}

years. The answer depends only on the death rates, not on how many people of each age the population contains.