Answer-first summary for fast verification
Answer: 6.20%
For a man aged 80, the probability of surviving the first year is: \[ 1 - q_{80} = 1 - 0.059403 = 0.940597 \] The probability of dying in the second year (ages 81 to 82) is then: \[ 0.940597 \times q_{81} = 0.940597 \times 0.065873 \approx 0.06196 = 6.20\% \] So the nearest answer is **C. 6.20%**.
Author: Manit Arora
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Q-702.1. Below is an extract (selected rows) from a mortality table:
| Age (Yrs) | Male | Female | ||||
|---|---|---|---|---|---|---|
| Probability | Probability | |||||
| Death w/n | Survival | Life Expect | Death w/n | Survival | Life Expect | |
| one year | one year | |||||
| 0 | 0.006519 | 1.000000 | 76.280000 | 0.005377 | 1.000000 | 81.05 |
| 1 | 0.000462 | 0.993481 | 75.780000 | 0.000379 | 0.994623 | 80.49 |
| 2 | 0.000291 | 0.993022 | 74.820000 | 0.000221 | 0.994246 | 79.52 |
| 3 | 0.000209 | 0.992733 | 73.840000 | 0.000162 | 0.994026 | 78.54 |
| ... | ... | ... | ... | ... | ... | ... |
| 79 | 0.053739 | 0.535041 | 8.730000 | 0.038920 | 0.664666 | 10.24 |
| 80 | 0.059403 | 0.506288 | 8.200000 | 0.043289 | 0.638797 | 9.64 |
| 81 | 0.065873 | 0.476213 | 7.680000 | 0.048356 | 0.611144 | 9.05 |
| 82 | 0.073082 | 0.444844 | 7.190000 | 0.054041 | 0.581592 | 8.48 |
| 83 | 0.081070 | 0.412334 | 6.720000 | 0.060384 | 0.550162 | 7.94 |
Which is nearest to the probability of a man aged 80 years old dying in the second year (between ages 81 and 82)?
A
0.39%
B
1.76%
C
6.20%
D
7.31%
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