Question 11.7.E.29: Work each problem. See Examples (1- 2 - 3 - 4 - 5 - 6). Male...
Work each problem. See Examples (1– 2 – 3 – 4 – 5 – 6).
Male Life Table The table is an abbreviated version of the 2010 period life table used by the Office of the Chief Actuary of the Social Security Administration. (The actual table includes every age, not just every tenth age.) Theoretically, this table follows a group of 100,000 males at birth and gives the number still alive at each age. Round answers to the questions that follow to the nearest thousandth.
| Exact Age | Number of Lives | Exact Age | Number of Lives |
| 0 | 100,000 | 60 | 86,010 |
| 10 | 99,155 | 70 | 73,355 |
| 20 | 98,731 | 80 | 49,939 |
| 30 | 97,450 | 90 | 17,152 |
| 40 | 95,889 | 100 | 873 |
| 50 | 92,820 | 110 | 2 |
Source: Office of the Actuary, Social Security Administration.
(a) What is the probability that a 40-year-old man will live 30 more years?
(b) What is the probability that a 40-year-old man will not live 30 more years?
(c) Consider a group of five 40-year-old men. What is the probability that exactly three of them survive to age 70? (Hint: The longevities of the individual men can be considered as independent trials.)
(d) Consider two 40-year-old men. What is the probability that at least one of them survives to age 70? (Hint: The complement of at least one is none.)
(a) A 40-yr-old man who lives 30 more yr would be 70 yr old.
Let E be the event “selected man will live to be 70”; then n(E) = 73,355. For this situation, the sample space S is the set of all 40-yr-old men, so n(S) = 95,889. Thus, the probability that a 40-yr-old man will live 30 more yr is
P(E) =\frac{ n(E)}{n(S)} = \frac{73,355}{95,889} ≈ 0.765.
(b) Using the notation and result from part (a), the probability that a 40-yr-old man will not live 30 more yr is
P(E′) = 1 – P(E) = 1 – 0.765 = 0.235.
(c) Use the notation and results from parts (a) and (b). In this binomial experiment, we call “a 40-yr-old man survives to age 70” a success.
Then,
p = P(E) = 0.765
and q = 1 – p = P(E′) = 0.235.
There are 5 independent trials and we need the probability of 3 successes, so n = 5 and r = 3. The probability that exactly 3 of the 40-yr-old men survive to age 70 is computed as follows.
\left(\begin{matrix} n\\r \end{matrix} \right) p^{r} q^{n-r}
=\left(\begin{matrix} 5\\3 \end{matrix} \right) (0.765)^{3}(0.235)^{2}
= 10(0.765)^{3}(0.235)^{2}
≈ 0.247
(d) Let F be the event “at least one man survives to age 70.” The easiest way to find P(F) is to first find the probability of the complementary event F′: “neither man survives to age 70.”
P(F′) = P(E′) • P(E′) = (0.235)² = 0.055225
Then,
P(F) = 1 – P(F′) = 1 – 0.055225 ≈ 0.945.