Revision as of 03:11, 20 January 2024 by Admin (Created page with "'''Answer: B''' <math>L_{A}=v^{T}-0.10 \bar{a}_{\bar{T}}=\left(1+\frac{10}{6}\right) v^{T}-\frac{10}{6}</math> <math>\operatorname{Var}\left(L_{A}\right)=\left(1+\frac{10}{6}\right)^{2} \operatorname{Var}\left(v^{T}\right)=0.455 \Rightarrow \operatorname{Var}\left(v^{T}\right)=0.06398</math> <math>L_{B}=2 v^{T}-0.16 \bar{a}_{T}=\left(2+\frac{16}{6}\right) v^{T}-\frac{16}{6}</math> <math>\operatorname{Var}\left(L_{B}\right)=\left(2+\frac{16}{6}\right)^{2} \operatornam...")
Exercise
ABy Admin
Jan 20'24
Answer
Answer: B
[math]L_{A}=v^{T}-0.10 \bar{a}_{\bar{T}}=\left(1+\frac{10}{6}\right) v^{T}-\frac{10}{6}[/math]
[math]\operatorname{Var}\left(L_{A}\right)=\left(1+\frac{10}{6}\right)^{2} \operatorname{Var}\left(v^{T}\right)=0.455 \Rightarrow \operatorname{Var}\left(v^{T}\right)=0.06398[/math]
[math]L_{B}=2 v^{T}-0.16 \bar{a}_{T}=\left(2+\frac{16}{6}\right) v^{T}-\frac{16}{6}[/math]
[math]\operatorname{Var}\left(L_{B}\right)=\left(2+\frac{16}{6}\right)^{2} \operatorname{Var}\left(v^{T}\right)=\left(2+\frac{16}{6}\right)^{2}(0.06398)=1.39[/math]
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