A father-son club has 4000 members, 2000 of which are age 20 and the other 2000 are age 45. In 25 years, the members of the club intend to hold a reunion.

You are given:

(i) All lives have independent future lifetimes.

(ii) Mortality follows the Standard Ultimate Life Table.

Using the normal approximation, without the continuity correction, calculate the [math]99^{\text {th }}[/math] percentile of the number of surviving members at the time of the reunion.

• 3810
• 3820
• 3830
• 3840
• 3850

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Answer: E

From the SULT, we have:

[[math]] \begin{aligned} { }_{25} p_{20} & =\frac{l_{45}}{l_{20}}=\frac{99,033.9}{100,000.0}=0.99034 \\ { }_{25} p_{45} & =\frac{l_{70}}{l_{45}}=\frac{91,082.4}{99,033.9}=0.91971 \end{aligned} [[/math]]

The expected number of survivors from the sons is 1980.68 with variance 19.133.

The expected number of survivors from fathers is 1839.42 with variance 147.687.

The total expected number of survivors is therefore 3820.10 .

The standard deviation of the total expected number of survivors is therefore

[[math]] \sqrt{19.133+147.687}=\sqrt{166.82}=12.916 [[/math]]

The [math]99^{\text {th }}[/math] percentile equals [math]3820.10+(2.326)(12.916)=3850[/math]

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.