Question

总体N（52.6，3的平方）中随机抽取一容量为36的样本，求样本均值X横落在50.8-53.8之间的概率

要有过程

Answer 1

P(50.8<X<53.8)
=P(-1.2<X-52<1.8)
=P(-1.2/[(6.3^2/36)^0.5]<(X-52)/[(6.3^2/36)^0.5]<1.8/[(6.3^2/36)^0.5])
=P(-1.1429<(X-52)/[(6.3^2/36)^0.5]<1.7143)
=f(1.7143)-[1-f(1.1429)] f(x)是标准正态分布函数
=0.9568-(1-0.8735)
=0.8303

Answer 2

0.9916