Question

#10) People may wonder ifthey will live to be at least 80 years old. Assume that a person borm in the United States will live to be for at least 80 years old. If you ran om. select 20 people born in the United States, find the estima there is a 63% chance them will live to be at least 80 years old. more than 10 of 8 pts. X-.10 n- 26 4.es rì) > 5-(24)(043): 2.157 of tne ep ran tre US. wil ive frit, ast, 20나cars old

Answer 1

$\dpi{100} \mathbf{Solution}$

$\dpi{100} \mathbf{given:}\;p=0.63,q=1-p=0.37,\;n=20$

$\dpi{100} np=20*0.63=12.6\geq 5$

$\dpi{100} nq=20*0.37=7.4\geq 5$

$\dpi{100} \therefore \mathbf{Binomial\; random\; variable \;is\; approximately\; normal}$

$\dpi{100} mean\;(\mu)=np=20*0.63=\mathbf{12.6}$

$\dpi{100} standard\;deviation\;(\sigma)=\sqrt{npq}=\sqrt{20*0.63*0.37}=\mathbf{\sqrt{4.662}}$

$\dpi{100} Let \;X\rightarrow People\; who\; live \;at \;least \;80\;years$

$\dpi{100} \mathbf{formula:}\;Z=\frac{X-\mu}{\sigma}$

$\dpi{100} P(more\;than\;10)\Rightarrow P(X>10)$

$\dpi{100} P(X>10)\Rightarrow P(X>10.5)\;(continuity\;correction)$

$\dpi{100} \Rightarrow P\left ( \frac{X-\mu}{\sigma}>\frac{10.5-12.6}{\sqrt{4.662}} \right )$

$\dpi{100} \Rightarrow P\left (Z>-0.97\right )$

.2 1-0.97 0

$\dpi{100} \Rightarrow \mathbf{{\color{red} 0.8340}}\;(\mathbf{from\;z-table})$