概率论与数理统计第三版课后答案习题4Word格式.docx
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第四章随机变量的数字特征
1.甲、乙两台自动车床，生产同一种零件，生产
1000
件产品所出的次品数分别用
，
表示，经过一段时间的考察，知
，的分布律如下：
1
2
3
p
0.7
0.1
0.5
0.3
0.2
试比较两台车床的优劣。
解：
因为E=00.7+10.1+20.1+30.1=0.6;
E=00.5+10.3+20.2=0.7。
故就平均来说，甲机床要优于乙机床。
2.连续型随机变量的概率密度为
kxa
x
（k,a
0）
f（x）
其它
又知E=0.75，求k,a之值。
f（x）dx
1,
即
kxadx
a
k
首先由密度函数性质知
；
xf（x）dx
0.75,
kxa1dx
0.75
又
E=0.75，即有
由上述两式可求得
k=3,a=2。
3.已知随机变量
的分布律为
-1
1/8
1/4
3/8
求E，E（3-2），E2，E（1-）2。
E
=（-1）（1/8）+0
（1/4）+2（3/8）+3（1/4）=11/8;
E
2=（-1）2
（1/8）+02（1/4）+22
（3/8）+32（1/4）=31/8;
（1/8）+（1-0）
E（1-）=（1-（-1））
（1/4）+（1-2）
（3/8）+（1-3）
（1/4）=17/8
或者，E（1-）2=E（1-2+
2）=1-（E2）+E2=17/8。
1e
4.若的概率密度为
|x|
。
求
（1）E，
（2）E2。
xe|x|dx
中因e-|x|为偶函数，x为奇函数，故
xe-|x|为奇函数，且积分区
（1）
间关于原点对称，该积分又绝对收敛，事实上
|x|f（x）dx
|x|e|x|dx
0xexdx
（2）
故E=0。
12
e
dx
dx（3）2!
2
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f（x）dx
0x
5.轮船横向摇摆的随机振幅
的概率密度为
f（x）Axe
x2
22
x0（0）
x0
求
（1）确定系数A；
（2）遇到大于其振幅均值的概率是多少？
f（x）dx1,即
Axe2
dx1,
A2
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