设X～N(10，22)，求：
(1)P{7﹤X≤15}；(2)常数d，使P{∣X-10∣﹤d}﹤0.9．
【正确答案】：(1)P{7﹤X≤15}=Φ[(15/10)/2]-Φ[(7/10)/2] =Φ(2.5)-Φ(-1.5) =Φ(2.5)-1+Φ(1.5) =0．9270 (2)P{∣X-10∣﹤d} =P{-d﹤X-10﹤d} = P{-d/2﹤(X-10)/2﹤d/2}=Φ(d/2)=Φ[-(d/2)]=2Φ(d/2)-1﹤0.9 ∴Φ(d/2)﹤0.95，反查正态分布表知d80=1.65 ∴d=3•3
