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(Created page with "'''Answer: E''' The distribution is binomial with 10,000 trials. <math display="block"> \begin{aligned} & \operatorname{Var}\left[L_{15}\right]=n p q=10,000\left({ }_{15} p_{50}\right)\left({ }_{15} q_{50}\right) \\ & { }_{15} p_{50}=e^{\left[-A(15)-\frac{B}{\ln C} c^{50}\left(c^{15}-1\right)\right]}=0.837445 \\ & { }_{15} q_{50}=1-{ }_{15} p_{50}=0.162555 \end{aligned} </math> <math>\operatorname{Var}\left[L_{15}\right]=10,000(0.837445)(0.162555)=1361.3</math>") |
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<math>\operatorname{Var}\left[L_{15}\right]=10,000(0.837445)(0.162555)=1361.3</math> | <math>\operatorname{Var}\left[L_{15}\right]=10,000(0.837445)(0.162555)=1361.3</math> | ||
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Latest revision as of 02:34, 18 January 2024
Answer: E
The distribution is binomial with 10,000 trials.
[[math]]
\begin{aligned}
& \operatorname{Var}\left[L_{15}\right]=n p q=10,000\left({ }_{15} p_{50}\right)\left({ }_{15} q_{50}\right) \\
& { }_{15} p_{50}=e^{\left[-A(15)-\frac{B}{\ln C} c^{50}\left(c^{15}-1\right)\right]}=0.837445 \\
& { }_{15} q_{50}=1-{ }_{15} p_{50}=0.162555
\end{aligned}
[[/math]]
[math]\operatorname{Var}\left[L_{15}\right]=10,000(0.837445)(0.162555)=1361.3[/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.
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.