设X~N(3,2^2),使d满足P{X>d}>=0.9,d至多是多少

Question

设X~N(3,2^2),使d满足P{X>d}>=0.9,d至多是多少

Answer 1

P(X>d) = 1-P(X<=d) = 1- P[(X-3)/2 <= (d-3)/2] =1-Φ[(d-3)/2].
令:P(X>d) >= 0.9,
即:1-Φ[(d-3)/2]>=0.9,
得:即:Φ[(d-3)/2]<=0.1.
查表得:(d-3)/2<=-1.28,
解得:d<=0.44
,即d至多为0.44.