设测量误差 sim N((0,5)^2) ,先进行100次测量,求误差的绝对值超过9.8的次数不-|||-小于3的概率-|||-A https:/img.zuoyebang.cc/zyb_88e7c818c80277bbb2c10eb6935ead71.jpg-dfrac (17)(2)(e)^-5-|||-B https:/img.zuoyebang.cc/zyb_88e7c818c80277bbb2c10eb6935ead71.jpg-dfrac (47)(2)(e)^-5-|||-C https:/img.zuoyebang.cc/zyb_88e7c818c80277bbb2c10eb6935ead71.jpg-dfrac (27)(2)(e)^-5-|||-D https:/img.zuoyebang.cc/zyb_88e7c818c80277bbb2c10eb6935ead71.jpg-dfrac (37)(2)(e)^-5
题目解答
答案
9.8)\text{,先将}X\text{标准化}: \\ &\begin{aligned}P(|X|>9.8)=P(X>9.8)+P(X<&-9.8)\end{aligned} \\ &\begin{aligned}P(X>9.8)=1-P(X\leqslant9.8)\end{aligned}\text{,}P(X\leqslant \\ &9.8)=\Phi(\frac{9.8-0}{5})=\Phi(1.96)\text{,查标准正态} \\ &\text{分布表得}\Phi(1.96)=0.975\text{,所以}P(X> \\ &9.8)=1-0.975=0.025。 \\ &\text{同理}P(X<-9.8)=\Phi(\frac{-9.8-0}{5})=\Phi(- \end{aligned}" data-width="473" data-height="638" data-size="58898" data-format="png" style="">
9.8)=0.025+0.025=\end{aligned} \\ &\text{0.05。} \\ 2.& \text{根据二项分布计算所求概率} \\ &\text{设Y表示100次测量中误差的绝对值超过} \\ &9.8\text{的次数,则}Y\sim B(100,0.05)。 \\ &P(Y\geqslant3)=1-P(Y=0)-P(Y=1) \\ &P(Y=2)_{0} \\ &\text{根据二项分布概率公式}P(Y=k)=C_{100}^k\times \\ &0.05^{k}\times\left(1-0.05\right)^{100-k}。 \\ &\begin{aligned}P(Y=0)=C_{100}^0\times0.05^0\times\left(1-0.05\right)^{100}=\end{aligned} \\ &0.95^{100} \\ &\begin{aligned}P(Y=1)=C_{100}^{1}\times0.05^{1}\times\left(1-0.05\right)^{99}=\end{aligned} \\ &100\times0.05\times0.95^{99} \\ &P(Y=2)=C_{100}^{2}\times0.05^{2}\times\left(1-0.05\right)^{98}= \\ &\frac{100\times99}{2}\times0.05^2\times0.95^{98}。 \\ &\text{计算}P(Y=0)\text{,}P(Y=1)\text{,}P(Y=2)\text{的} \\ &\text{值:} \\ &P(Y=0)\approx0.00592\textit{,}P(Y=1)\approx \\ &0.03116\text{,}P(Y=2)\approx0.08118\text{。}\\ &P(Y\ge3)=1-(0.00592+0.03116+\\ &0.08118)=1-0.11826=1-\frac{27}{2}e^{-5}\\ &\text{所以本题应该选C。} \end{aligned}" data-width="417" data-height="767" data-size="82695" data-format="png" style="">