X~N(3,4),则[(x-3)/2]~N(0,1),所以P(2＜X≤5)=P(-0.5＜[(X-3)/2]≤1)=1-P(x≤-0.5)-P(X＞1)=1-0.3085-0.1587=0.5328,P(-4＜X≤10)=P(-3.5＜[(X-3)/2]≤3.5)≈1,P(｜X｜＞2)=P(X＞2) P(X＜-2)=P(X＞-0.5) P(X＜-2.5)=0.6915 0.0062=0.6977,P(X＞3)=P([(X-3)/2]＞0)=0.5,关于第二问,必然是P(X＞c)=0.5,因为知道P([(X-3)/2]＞0)=0.5,所以c=0×2 3=3 X~N(3,4),则[(x-3)/2]~N(0,1),所以P(2＜X≤5)=P(-0.5＜[(X-3)/2]≤1)=1-P(x≤-0.5)-P(X＞1)=1-0.3085-0.1587=0.5328,P(-4＜X≤10)=P(-3.5＜[(X-3)/2]≤3.5)≈1,P(｜X｜＞2)=P(X＞2) P(X＜-2)=P(X＞-0.5) P(X＜-2.5)=0.6915 0.0062=0.6977,P(X＞3)=P([(X-3)/2]＞0)=0.5,关于第二问,必然是P(X＞c)=0.5,因为知道P([(X-3)/2]＞0)=0.5,所以c=0×2 3=3