设X~N(3,4)(1)求P｛2＜X≤5｝,P｛-4＜X≤10｝,P｛｜X｜＞2｝,P｛X＞3｝(2)确定c使得P｛X＞c｝=P｛X≤c｝

⑴X~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

⑵P(X＞c)=0.5，因为知道P([(X-3)/2]＞0)=0.5，所以c=0×2+3=3