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在总体N(12,4)中随机抽一容量为5的样本X1,X2,X3,X4,X5.
(1)求样本均值与总体均值之差的绝对值大于1的概率.
(2)求概率P{max{ X1,X2,X3,X4,X5}>15};P{min{ X1,X2,X3,X4,X5}
人气:392 ℃ 时间:2019-10-19 23:06:48
解答
1、样本均值服从N(12,0.8)
P(|样本均值-12|>1)=P(|样本均值-12|/根号0.8>根号5/2)=2F(1.118)-1
=0.7698
2、P{max{ X1,X2,X3,X4,X5}>15}
=1-P{max{ X1,X2,X3,X4,X5}≤15}
=1-[P(X≤15)]^5
=1-[P(X-12)/2≤1.5)]^5
=1-F(1.5)^5
1-0.9332^5=0.3023;
P{min{ X1,X2,X3,X4,X5}<10}=1-P{min{ X1,X2,X3,X4,X5}≥10}.
=1-[P(X≥10)]^5=1-[1-P(X-12)/2<-1)]^5=1-F(1)^5
=1-(0.8413)^5=0.5786
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