题目
设X1,X2是随机变量,且 _(1)sim N(0,1), _(2)sim N((5.2)^2), 若_(i)=P(-2leqslant (X)_(i)leqslant 2)(i=1,2), 则-|||-() ←-|||-bigcirc A.无法确定-|||-bigcirc B. _(1)=(p)_(2)-|||-bigcirc C. _(1)lt (p)_(2)-|||-D. _(1)gt (p)_(2)
题目解答
答案
解析
步骤 1：计算 ${P}_{1}$
${P}_{1}=P\{ -2\leqslant {X}_{1}\leqslant 2\} =\phi (2)-\phi (-2)$
$=2\phi (2)-1$
步骤 2：计算 ${P}_{2}$
${P}_{2}=P\{ -2\leqslant {X}_{2}\leqslant 2\}$
$=P\{ \dfrac {-2-5}{2}\leqslant \dfrac {{X}_{2}-5}{2}\leqslant \dfrac {2-5}{2}\}$
$=\phi (-\dfrac {3}{2})-\phi (-\dfrac {7}{2})$
$=2\phi (-\dfrac {3}{2})-1$
步骤 3：比较 ${P}_{1}$ 和 ${P}_{2}$
由正态分布的性质可知，$\phi (2) > \phi (-\dfrac {3}{2})$，因此 ${P}_{1} > {P}_{2}$。
${P}_{1}=P\{ -2\leqslant {X}_{1}\leqslant 2\} =\phi (2)-\phi (-2)$
$=2\phi (2)-1$
步骤 2：计算 ${P}_{2}$
${P}_{2}=P\{ -2\leqslant {X}_{2}\leqslant 2\}$
$=P\{ \dfrac {-2-5}{2}\leqslant \dfrac {{X}_{2}-5}{2}\leqslant \dfrac {2-5}{2}\}$
$=\phi (-\dfrac {3}{2})-\phi (-\dfrac {7}{2})$
$=2\phi (-\dfrac {3}{2})-1$
步骤 3：比较 ${P}_{1}$ 和 ${P}_{2}$
由正态分布的性质可知，$\phi (2) > \phi (-\dfrac {3}{2})$，因此 ${P}_{1} > {P}_{2}$。