概率论与数理统计习题二答案 下载本文

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【解】当x<0时F(x)=0

当0≤x<1时F(x)?x?x??f(t)dt??0??f(t)dt??f(t)dt

0xx2 ??tdt?

02当1≤x<2时F(x)??x??f(t)dt

??

0??1f(t)dt??f(t)dt??f(t)dt01x11x??tdt??(2?t)dt01x23??2x??222x2???2x?12

当x≥2时F(x)??x??f(t)dt?1

?0,?2?x,?2故 F(x)??2??x?2x?1,?2??1,26.设随机变量X的密度函数为

(1) f(x)=ae?|x|,λ>0;

x?00?x?1

1?x?2x?2?bx,0?x?1,?11?x?2, (2) f(x)=?2,?x?0,其他.试确定常数a,b,并求其分布函数F(x). 【解】(1) 由

????f(x)dx?1知1??ae?????|x|dx?2a?e??xdx?0?2a?

故 a??2

????xe,x?0??2即密度函数为 f(x)??

??e?xx?0??2当x≤0时F(x)??x??f(x)dx??1e?xdx?e?x ??22x?.

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当x>0时F(x)??x??f(x)dx??1??xe 20?2??edx???xx?20e??xdx

?1?故其分布函数

?1??x1?e,x?0??2F(x)??

1?e?x,x?0??2(2) 由1?????f(x)dx??bxdx??01211b1dx?? 2x22得 b=1

即X的密度函数为

0?x?1?x,?1?f(x)??2,1?x?2

?x其他??0,当x≤0时F(x)=0 当0

0x ??x0x2xdx?

2x??当1≤x<2时F(x)? ??f(x)dx??0dx??xdx????001x11dx x231? 2x当x≥2时F(x)=1 故其分布函数为

?0,?2?x,?F(x)??2?3?1,?2x?1,?27.求标准正态分布的上?分位点, (1)?=0.01,求z?; (2)?=0.003,求z?,z?/2. 【解】(1) P(X?z?)?0.01

.

x?00?x?1

1?x?2x?2.

即 1??(z?)?0.01 即 ?(z?)?0.09 故 z??2.33 (2) 由P(X?z?)?0.003得

1??(z?)?0.003

即 ?(z?)?0.997 查表得 z??2.75 由P(X?z?/2)?0.0015得

1??(z?/2)?0.0015

即 ?(z?/2)?0.9985 查表得 z?/2?2.96 28.设随机变量X的分布律为 X Pk ?2 ?1 0 1 3 1/5 1/6 1/5 1/15 11/30 求Y=X2的分布律.

【解】Y可取的值为0,1,4,9

P(Y?0)?P(X?0)?15117??61530

P(Y?1)?P(X??1)?P(X?1)?1511P(Y?9)?P(X?3)?30P(Y?4)?P(X??2)?故Y的分布律为 Y Pk 0 1 4 9 1/5 7/30 1/5 11/30 29.设P{X=k}=(

1k

), k=1,2,…,令 2?1,当X取偶数时Y??

??1,当X取奇数时..

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求随机变量X的函数Y的分布律. 【解】P(Y?1)?P(X?2)?P(X?4)??P(X?2k)?

111?()2?()4??()2k?222

111?()/(1?)?443

2P(Y??1)?1?P(Y?1)?

330.设X~N(0,1).

(1) 求Y=eX的概率密度; (2) 求Y=2X2+1的概率密度; (3) 求Y=|X|的概率密度.

【解】(1) 当y≤0时,FY(y)?P(Y?y)?0

x当y>0时,FY(y)?P(Y?y)?P(e?y)?P(X?lny)

??lny??fX(x)dx

故 fY(y)?(2)P(Y?2X?1?1)?1

2dFY(y)111?ln2y/2?fx(lny)?e,y?0 dyyy2π当y≤1时FY(y)?P(Y?y)?0

2当y>1时FY(y)?P(Y?y)?P(2X?1?y)

?P?X? ???2?y?1?y?1?P??X????2?2?y?1? ??2??(y?1)/2?(y?1)/2fX(x)dx

?y?1??y?1???fX??? ???????2?2?????d1FY(y)?故 fY(y)?dy4 ?(3) P(Y?0)?1

2??fXy?1??1221?(y?1)/4e,y?1

y?12π当y≤0时FY(y)?P(Y?y)?0

.

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当y>0时FY(y)?P(|X|?y)?P(?y?X?y) ?故fY(y)??y?yfX(x)dx

dFY(y)?fX(y)?fX(?y) dy2?y2/2e,y?0 2π?31.设随机变量X~U(0,1),试求:

(1) Y=eX的分布函数及密度函数; (2) Z=?2lnX的分布函数及密度函数. 【解】(1) P(0?X?1)?1

) 1故 P(1?Y?e?e?当y?1时FY(y)?P(Y?y)?0

X当1

X??lny0dx?lny

X当y≥e时FY(y)?P(e?y)?1

即分布函数

y?1?0,?FY(y)??lny,1?y?e

?1,y?e?故Y的密度函数为

?11?y?e?fY(y)??y,

?0,其他?(2) 由P(0

P(Z?0)?1

当z≤0时,FZ(z)?P(Z?z)?0

当z>0时,FZ(z)?P(Z?z)?P(?2lnX?z)

?P(lnX??)?P(X?ez2?z/2)

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