例 2.3.4 (重复测量中误差超限次数)

已知测量误差 $X\sim N\left(0,{10}^2\right)$. 现独立的重复进行 100 次测量, 求误差的绝对值超过 19.6 的次数不少于 3 的概率.

解

先求一次测量误差的绝对值超过 19.6 的概率.

$$\begin{array}{rl}P\left(|X|>19.6\right)& =1-P\left(\frac{|X|}{10}\le 1.96\right)\\ & =1-\left[\Phi (1.96)-\Phi (-1.96)\right]\\ & =2-2\Phi (1.96)\\ & =2-2\times \mathrm{pnorm}(1.96)\\ & \approx \mathrm{0.05.}\end{array}$$

所以 记 $Y$ 为 100 次测量中误差的绝对值超过 19.6 的次数, 则 $Y\sim B\left(100,0.05\right)$,

$$\begin{array}{rl}P\left(Y\ge 3\right)& =1-P\left(Y\le 2\right)\\ & =1-\mathrm{pbinom}(2,100,0.05)\\ & \approx \mathrm{0.881737.}\end{array}$$

若利用泊松逼近, 则 $\lambda =100\times 0.05=5$, 得

$$\begin{array}{rl}P\left(Y\ge 3\right)& =1-P\left(Y\le 2\right)\\ & \approx 1-\mathrm{ppois}(2,5)\\ & \approx \mathrm{0.875348.}\end{array}$$