武汉大学2013-2014学年《概率论与数理统计》期末考试试卷 (B)

武汉大学2013-2014学年《概率论与数理统计》期末考试试卷 (B)
武汉大学2013-2014学年《概率论与数理统计》期末考试试卷 (B)

武汉大学2013-2014学年《概率论与数理统计》

期末考试试卷 (B)

一、填空题(每小题4分,共32分).

1.设 A 、B 为随机事件, P (A ) = 0.3, P (B ) = 0.4, 若 P (A |B ) =0.5, 则 P (A ?B ) = _______; 若 A 与 B 相互独立, 则 P (A ?B ) = _________.

2.设随机变量 X 在区间 [0, 10] 上服从均匀分布, 则 P { 1 < X < 6} = ______________.

3.设随机变量 X 的分布函数为,4

,1 42 ,7.021 ,2.01 ,0 )(????

??

?≥<≤<≤--<=x x x x x F 则 X 的分布律为 ___________________________ . 4.若离散型随机变量 X 的分布律为

则常数 a = _________; 又 Y = 2X + 3, 则 P {Y > 5} = _________ . 5.设随机变量 X 服从二项分布 b (100, 0.2), 则 E (X ) = ________, D (X ) = ___________.

6.设随机变量 X ~ N (0, 1), Y ~ N (1, 3), 且X 和 Y 相互独立, 则D (3X +2Y ) =

_________.

7.设随机变量 X 的数学期望 E (X ) = μ, 方差 D (X ) = σ 2, 则由切比雪夫不等式有 P {|X - μ | <2σ } ≥ _________________.

8.从正态总体 N (μ, σ 2)(σ 未知) 随机抽取的容量为 25的简单随机样本, 测得样本均值5=x ,样本的标准差s = 0.1,则未知参数 μ 的置信度为0.95的置

信区间是 ____________________________. (用抽样分布的上侧分位点表示).

二、选择题(只有一个正确答案,每小题3分,共18分)

1.设随机事件A 与B 互不相容,且0)(,0)(>>B P A P ,则 ( ).

(A) )(1)(B P A P -= (B) )()()(B P A P AB P = (C) 1)(=B A P (D) 1)(=AB P

2.设随机变量 X 的概率密度为)(x f X , 则随机变量X Y 2-=的概率密度为

)(y f Y 为 ( ).

(A) )2-(2y f X (B) )2(y f X - (C) )2(21y f X - (D) )2

(21y

f X --

3.设随机变量 X 的概率密度为)(e

21)(4

)2(2

+∞<<-∞=

+-

x x f x π

,且

b aX Y +=)1,0(~N ,则下列各组数中应取 ( ).

(A)1,21==

b a (B) 2,22==b a (C) 1,21-==

b a (D) 2,2

2-==b a 4. 设两个相互独立的随机变量 X 和 Y 分别服从正态分布 ),(211σμN 和

),(2

22σμN , 则Y X Z +=也服从正态分布,且 ( ).

),(~ )A (2

2211σσμ+N Z ),(~ )B (2121σσμμ+N Z ),(~ )C (222121σσμμ+N Z ),(~ )D (222121σσμμ++N Z

5.对任意两个相互独立的随机变量 X 和 Y , 下列选项中不成立的是 ( ). (A) D (X + Y ) = D (X ) + D (Y ) (B) E (X + Y ) = E (X ) + E (Y )

(C) D (XY ) = D (X )D (Y ) (D) E (XY ) = E (X )E (Y )

6.设 X 1, X 2为来自总体 N (μ, 1) 的一个简单随机样本, 则下列估计量中μ 的无

偏估计量中最有效的是 ( ).

(A) 212121

X X +

=μ (B) 213231

X X +=μ (C) 21434

1

X X +

=μ (D) 215

352

X X +=μ 三、解答(本题 8 分)一个袋中共有10个球,其中黑球3个,白球7个,先从袋中先后任取一球(不放回)(1) 求第二次取到黑球的概率; (2) 若已知第二次取到的是黑球,试求第一次也取到黑球的概率?

四、解答(本题8分)设连续型随机变量 X 的概率密度为,

其他???≤≤+= ,0 2

0,1)(x ax x f 求: (1) 常数 a 的值; (2) 随机变量 X 的分布函数 F (x ); (3) }.21{<

??

?<<=-其他,

0,

,0,e ),(x y y x f x 求: (1) 求 X , Y 的边缘概率密度 f X (x ), f Y (y ), 并判断 X 与 Y 是否相互独立(说明原因)? (2) 求 P { X + Y ≤ 1}.

六、解答(本题8分)已知随机变量 X 分布律为

求 E (X ), D (X ).

七、(本题6分)对敌人的防御阵地进行100次轰炸,每次轰炸命中目标的炸弹

数目是一个随机变量,七期望值是2,方差是1.69。求在100次轰炸中有180颗到220颗炸弹命中目标的概率。其中9382.0)54.1(=Φ.

八、(10分) 设总体 X 的概率密度为,其他?

??<<= ,0 1

0 ,)(1-x x x f θθ其中 θ >0 是未知

参数, X 1, X 2, …, X n 为来自总体的一个简单随机样本,x 1, x 2, …, x n 为样本值, 求 θ 的矩估计量和极大似然估计量.

参考答案: 一、填空题 1. 0.5 ;0.58 2. 3/5 3.

4. 0.2 ;0.5

5. 20 ;16 6. 21 7. 3/4 8. ))24(5

1.05,)24(51.05(025.0025.0t t +-

二、选择题

1. D

2. C

3. B

4. D

5. C

6. A

三、解答题

解:设A 事件表示“第二次取到黑球,B 1事件表示“第一次取到黑球”,B 2事件表示“第一次取到白球”, (1) 第二次取到黑球的概率:

)()()()()(2211B P B A P B P B A P A P +=

3.010

79310392=?+?=

(2) 若已知第二次取到的是黑球,试求第一次也取到黑球的概率: 9

2

3.0103

92)()()()(111=?==A P B P B A P A B P

四、解答题 解:(1) 22d )1(d )(120+=+==??∞∞-a x ax x x f 2

1-=∴a

(2) ?

-=x

t t f x F d )()(

0d 0d )()(0===≤??

∞-∞-x

x t t t f x F x 时,当

x x t t t t t f x F x x

x

+=++=

=<

00

41-d 121-0d d )()(20)(时,当 10d d 12

1

-0d d )()(22200=+++==

≥????

∞-∞

-x x

t t t t t t f x F x )(时,当

所以

?∞

-=x t t f x F d )()(=??

???

≥<<+≤2,120,4

1-0,02

x x x x x

(3) 4

1

)141(1)1()2(}21{=+--=-=<

五、解答题 (1) ??

???+∞

<≤===

??

-其它,00,e d e d ),()(0

--x x y y y x f x f x x x X ??

???+∞

<≤===??

+∞

-∞

-其它,00,e d e d ),()(-y x x y x f y f y

y x Y 因为 ),()()(y x f y f x f Y X ≠?,所以X 与Y 不是相互独立的. (2) 2

21

2

11

1

11-21

e -1e

2e 1d e e

d e d }1{)()(-

-

----=-+=-==≤+??

?

y x y Y X P y y

y

y

x

六、解答题

1.035.023.001.01)(?+?+?+?-=X E =1.2

1.035.023.001.0)1()(22222?+?+?+?-=X E =3 222

2.13])([)()(-=-=X E X E X D =1.56 七、解答题

解:设X i 为第i 轰炸命中目标的炸弹数目}200180{100

1≤≤∑=i i X P

}69

.1100210020069

.11002

10069

.11002100180{100

1

?-≤

?-≤

?-=∑=i i

X

P

4382.02

1

1)54.1()]54.1(1[)0()54.1()0(=+

-=--=--≈ΦΦΦΦΦ

八、解答题

解:(1) 矩估计法

1

d )(1

1-1+=

==?θθθμθx x x X E

1

11μμθ-=

∴ ∑===n

i i X n X A 111 所以 θ的矩估计量∧θX X -=1 (2) 最大似然法

似然函数 1-1

θθi n

i x L =∏= ,10<

1

-1

θθi n i x L =∏=1-1

θθi n

i n

x =∏=

∑=+=n

i i x n L 1

ln 1-ln ln )(θθ

∑=+=n

i i x n L 1

ln d ln d θθ 令0d ln d =θL 得θ的最大似然估计值 ∧

θ∑=-

=n

i i

x

n

1

ln

θ的最大似然估计量 ∧

θ∑=-

=n

i i

X

n

1

ln

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