内容发布更新时间 : 2025/3/19 6:01:34星期一 下面是文章的全部内容请认真阅读。
??(0.04)??1??(1.85)? ?0.5160?(1?097678) =0.4838;
(3)P(?2.80????1.21)??(?1.21)??(?2.80)?1??(1.21)??1??(2.80)?
??(2.80)??(1.21)?0.9974?0.8869?0.1105 3.16设随机变量ξ服从正态分布N(108，9)， （1）求P（101.1<ξ<117.6);
(2 ) 求常数α,使P(ξ<α)=0.90;
(3 ) 求常数α,使P(｜ξ-α｜>α)=0.01。 解: (1) P（101.1<ξ<117.6) = P(?2.3???1083?3.2)
??(3.2)??(?2.3)??(3.2)??1??(2.3)?
?0.999313?1?0.989276?0.988589;
a?108????108P???a??P????0.90,.所以33??(2)
a?108?1.28,即a?111.84;3?3?P???a?a??P???a?a??P???a??a????1082a?108?
?P???2a??P???0??P???33?????108??2a?108??P???36??1?????0.01,33????2a?108?2a?108?所以???2.33即a?57.5 ??0.99,查表得33??3.17某种电池的寿命?服从正态Na,?2分布,其中a?300(小时),??35(小时)(1)求电池寿命在250小时以上的概率;(2)求x,使寿命在a?x与a?x之间的概率不小于0.9??300解:(1)P???250??P(??1.43)35???300??P??1.43????1.43??0.9236;?35?x??300x??2?P?a?x???a?x??P??????3535??35?x??x??x???????????2????1?0.9,即?35??35??35?x?x?????0.95,所以?1.65,即x?57.753535????
3.18设??x?为N?0,1?分布的分布函数,证明当x?0有1e2x?x22?11?1???x??ex2x12x?x22?11???3??xx?12x证:1???x??1?1?e2x2??x??e?y22dy?y22?x?e?y22dy
?x2211??x2x?x1ey2?dy21x2?11?1e??3??2x2x?xx?2??x3?y2edy,所以y421?x211?x2e??1???x??ex2x2x?11???3??xx???，??的联合分布函数为F?x,y?,F?x,y?表示下列概率:3。19设二维随机变量?1?P?a???b,c???d?;?2?P?a???b,??y?;?3?P???a,??y?;?4?P???x,?????;?5?P?????,?????解：?1?P?a???b,c???d??F?b,d??F?b,c??F?a,d??F?a,c?;?2?P?a???b,??y??P???b,??y??P???a,??y??F?b?0,y??F?a,y?;?3?P???a,??y??P???a,??y??P???a,??y??F?a?0,y??F?a,y?;?4?P???x,??????F?x,????F??x?;?5?P?????,??????0
3．20 设二维随机变量（?,?）的联合函数为F（x,y），用它表示（?,?）落在区域D（如
下图所示）内的概率：
y
b b b b
543D
2
b
1O
a
1a a
23a4
解： </