解: 1z(nt-μ)/o=(n105-12)0.32=1.5221 R(10)1-Φ(-1.5221)=0.9360 2.(105)= Φ(-1.5220.32×10 4.2/106h 1-Φ(-1.522) 3.R(tb.95=1-(z=0.95Φ(z)=0.05 查表得: 1.64485 lnto9s=12+(-1.64485)×0.32=1147365 ∴to95=e147365=96148h解： 1.z= (㏑t−)/=(㏑105−12)/0.32=−1.5221 R(105 )=1−(−.)=0.9360 (−.)/0.32 105 1 −(−.) 3. R(t0.95)=1−(z)=0.95 (z)=0.05 查表得：z= −.64485 ㏑ t0.95=12+(−.64485)0.32=11.47365 ∴t0.95=e11.47365=96148h 2. (105 )= =4.2/106h