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<math>\Rightarrow 95 \% \mathrm{CI}</math> is approximately <math>(0.8 \pm 1.96(0.01265))=(0.775,0.825)</math> | <math>\Rightarrow 95 \% \mathrm{CI}</math> is approximately <math>(0.8 \pm 1.96(0.01265))=(0.775,0.825)</math> | ||
{{soacopyright|2024}} | {{soacopyright|2024}} | ||
Latest revision as of 02:36, 18 January 2024
Answer: D
[math]\hat{S}(1)=0.8[/math]
[math]\operatorname{Var}[S(1)]=\frac{S(1)(1-S(1)}{n} \approx \frac{(0.8)(0.2)}{1000}=0.01265^{2}[/math]
[math]\Rightarrow 95 \% \mathrm{CI}[/math] is approximately [math](0.8 \pm 1.96(0.01265))=(0.775,0.825)[/math]
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.