X~N(0,4)-->X/2~N(0,1)(设标准差为2) P(X>3.92)-->P(X/2>1.96)=1-Φ(1.96)=1-0.975=0.025-->P(∣X/2∣>1.96)=0.05 10次独立重复测量=10次独立伯努利实验 至少有二次测量误差的绝对值大约3.92的概率 =1-(1-0.05)^10-C(10,1)*(1-0.05)^9*0.05 =1-0.5987-10*0.6302*0.05=1-0.5987-0.3151=0.0862 X~N(0,4)-->X/2~N(0,1)(设标准差为2) P(X>3.92)-->P(X/2>1.96)=1-Φ(1.96)=1-0.975=0.025-->P(∣X/2∣>1.96)=0.05 10次独立重复测量=10次独立伯努利实验 至少有二次测量误差的绝对值大约3.92的概率 =1-(1-0.05)^10-C(10,1)*(1-0.05)^9*0.05 =1-0.5987-10*0.6302*0.05=1-0.5987-0.3151=0.0862