浙江 2017 · 数学 q08

题面

8．（5 分）已知随机变量$\xi_{\mathrm{i}}$满足$P ( \xi_{i} = 1 ) = p_{i}$，$P ( \xi_{i} = 0 ) = 1 - p_{i}$，i=1，2．若0 <$$p_{1} < p_{2} < \frac{1}{2}，则（ ）A$\begin{array}{r}{E ( \xi_{1} ) \ < E ( \xi_{2} ) , \ D ( \xi_{1} ) \ < D ( \xi_{2} ) B . E ( \xi_{1} ) \ < E ( \xi_{2} ) , \ D ( \xi_{1} ) \ > D ( \xi_{2} )} \end{array}$C．E$( \xi_{1} ) > E ( \xi_{2} )$），D$( \xi_{1} ) < D \ ( \xi_{2} )$D$E \left( \xi_{1} \right) > E \ \left( \xi_{2} \right)$），D$( \xi_{1} ) > D \ ( \xi_{2} )$

答案

A

解析

分析

由已知得$0 < p_{1} < p_{2} < \frac{1}{2} , \frac{1}{2} < 1 - p_{2} < 1 - p_{1} < 1$，求出$E ( \xi_{1} ) = \pmb{\rho}_{1} , E ( \xi_{2} )$

=p₂，从而求出$D ( \xi_{1} ) , D ( \xi_{2} )$，由此能求出结果

解答

解：∵随机变量ξ_i满足P ( \xi_{i} = 1 ) = p_{i} , P ( \xi_{i} = 0 ) = 1 - p_{i} , i = 1 , 2 , \ldots ,$$0 {<}{p}_{1}{<}{p}_{2}{<} \frac{1}{2} ,$$\therefore{\frac{1}{2}} < 1 - {p}_{2} < 1 - {p}_{1} < 1E$( \xi_{1} ) \ = 1 \times{p}_{1}{+} 0 \times \ ( \ 1 \ {-} \ {p}_{1} ) \ {=}{p}_{1}$E$( \xi_{2} ) \ = 1 \times{p}_{2} + 0 \times \ ( \mathrm{1} - {p}_{2} ) \ {= p}_{2}$D$( \xi_{1} ) = ( {1} - {p}_{1} ) {\xi}^{2}{p}_{1} + {\xi} ( {0} - {p}_{1} ) {\xi}^{2} ( {1} - {p}_{1} ) =_{{p}_{1} - {p}_{1}}{\xi}^{2} ,$D$( \xi_{2} ) = ( {1} - {p}_{2} )^{2}{p}_{2} + ( {0} - {p}_{2} )^{2} ( {1} - {p}_{2} ) = {p}_{2} - {p}_{2}{\xi}^{2} ,$D( \xi_{1} ) - {D} ( \xi_{2} ) = {p}_{1} - {p}_{1}{}^{2} - ( \mathrm{\vec{}{\ p}}_{2} - {p}_{2}{}^{2} ) = ( {p}_{2} - {p}_{1} ) ( {p}_{1} + {p}_{2} - 1 ) < 0 ,$$\therefore E ( \xi_{1} ) < E ( \xi_{2} ) , D ( \xi_{1} ) < D ( \xi_{2} )

故选：A