例 2.3.3 (正态分布的概率计算)

设 $X\sim N\left(-1,4\right)$, 试求

• $P\left(-5\le X<1\right)$,
• $P\left(-2\le X\le 2\right)$
• $P\left(\left|X\right|<1\right)$,
• $P\left(\left|X\right|\ge \frac{3}{2}\right).$

解

由题设, $\mu =-1,\sigma =2$, 由 (2.3.12), (2.3.13) 和 $\mathrm{R}$ 软件中的函数 pnorm,

$$\begin{array}{rl}P\left(-5\le X<1\right)& =\Phi \left(\frac{1-(-1)}{2}\right)-\Phi \left(\frac{-5-(-1)}{2}\right)\\ & =\Phi (1)-\Phi (-2)\\ & =\Phi (1)+\Phi (2)-1\\ & =\mathrm{pnorm}(1)+\mathrm{pnorm}(2)-1\\ & =0.8185946\end{array}$$ $$\begin{array}{rl}P\left(-2\le X\le 2\right)& =\Phi \left(\frac{2-(-1)}{2}\right)-\Phi \left(\frac{-2-(-1)}{2}\right)\\ & =\Phi \left(\frac{3}{2}\right)-\Phi \left(-\frac{1}{2}\right)\\ & =\mathrm{pnorm}\left(\frac{3}{2}\right)-\mathrm{pnorm}\left(-\frac{1}{2}\right)\\ & =0.6246553\end{array}$$ $$\begin{array}{rl}P(|X|<1)& =\Phi \left(\frac{1-(-1)}{2}\right)-\Phi \left(\frac{-1-(-1)}{2}\right)\\ & =\Phi (1)-\Phi (0)\\ & =\mathrm{pnorm}(1)-\mathrm{pnorm}(0)\\ & =\mathrm{0.3413447.}\end{array}$$ $$\begin{array}{rl}P\left(|X|\ge \frac{3}{2}\right)& =P\left(X\ge \frac{3}{2}\right)+P\left(X\le -\frac{3}{2}\right)\\ & =1-\Phi \left(\frac{\frac{3}{2}-(-1)}{2}\right)+\Phi \left(\frac{-\frac{3}{2}-(-1)}{2}\right)\\ & =1-\Phi \left(\frac{5}{4}\right)+\Phi \left(-\frac{1}{4}\right)\\ & =1-\mathrm{pnorm}\left(\frac{5}{4}\right)+\mathrm{pnorm}\left(-\frac{1}{4}\right)\\ & =\mathrm{0.5069434.}\end{array}$$