概率论与数理统计统计课后习题答案-总主编-邹庭荣-主编-程述汉-舒兴明-第四章
第四章习题解答
1.设随机变量X~B(30,
1 A.;
61),则E(X)=( D ). 6 B.
525; C.; 66 D.5.
1E(X)?np?30??5
62.已知随机变量X和Y相互独立,且它们分别在区间[-1,3]和[2,4]上服从均匀分布,则E(XY)=( A ).
A. 3; B. 6; C. 10; E(X)?1E(Y)?3
D. 12.
因为随机变量X和Y相互独立所以E(XY)?E(X)E(Y)?3
3.设X表示10次独立重复射击命中目标的次数,每次射中目标的概率为0.4,则X2的数学期望E(X 2)=____18.4______.
X:B(10,0.4)E(X)?4D(X)?2.4
E(X2)?(E(X))2?D(X)?18.4
234.某射手有3发子弹,射一次命中的概率为,如果命中了就停止射击,否则一直射到子弹用尽.设表示X耗用的子弹数.求E(X).
解:
X P E(X)?1 2/3 2 2/9 3 1/9 22113??2??3? 39995.设X的概率密度函数为
0?x?1?x,?f(x)??2?x,1?x?2
?0,其它?求E(X) ,E(X2). 解:E(X)??????xf(x)dx??x2dx??x(2?x)dx?1,
0112E(X2)??
????x2f(x)dx??x3dx??x2(2?x)dx?01127. 61
1
14E(X2)?[E(X)]2?D(X)?1??.
33E(X4)??????x4f(x)dx??201416xdx? 25D(X2)?E(X4)?[E(X2)]2?164264 ?()?534510.设随机变量X的分布律为
X P 求 D(X).
-1 1/5 0 1/2 1 1/5 2 1/10 1111 解:D(X)?E(X2)?(E(X))2,E(X)??1??0?1??2??,
551051114E(X2)?(?1)2??0?12??22??,
55105D(X)?E(X2)?(E(X))2?4119. ??52525111.设随机变量X的概率密度函数为f(x)?e?|x|,求D(X ).
2解:E(X)??E(X)??2????xf(x)dx??2????1?xxedx?0, 2??????xf(x)dx?2?012?xxedx?2, 2D(X)?E(X2)?(E(X))2?2.
12.设随机变量X,Y相互独立,其概率密度函数分别为
0?x?1?x,?e?y,y?0?fX(x)??2?x,1?x?2 fY(y)??
?0,其它?0,其它?求D(X ),D(Y ),D(X-Y ).
解:由本章习题5知E(X)?1,E(X2)?D(X)?E(X2)?(E(X))2?1. 67,于是有 6由Y:E(1)知E(X)?D(X)?1.
1
由于随机变量X,Y相互独立,所以
7. 613.设D(X)=1,D(Y)=4,相关系数?XY?0.5,则cov(X,Y)=___1____. D(X?Y)?D(X)?D(Y)?cov(X,Y)=?XYD(X)D(Y)?1
14.设二维随机变量(X, Y )的联合密度函数为
???1?sin(x?y),0?x?,0?y?f(x,y)??222
?0,其它?求cov(X,Y ),?XY. 解:E(X)??E(X)??2??????????????1??xf(x,y)dxdy??2?2xsin(x?y)dxdy?,
2004??????1?xf(x,y)dxdy??2?2x2sin(x?y)dxdy
20021?1???2(cosx+sinx)dx??2??2, 2082D(X)?E(X2)?[E(X)]2?12????2. 162由对称性 E(Y)?E(X)??4, D(Y)?D(X)?12????2. 162E(XY)???? ?1???2??2?2(xy)sin(x?y)dxdy?,0022 ?()2=-0.0461,24cov(x,y)??2?21??[?()](?2??2)=-0.2454.
24162D(X)D(Y)???????(xy)f(x,y)dxdycov(X,Y )=E(XY)?E(X)E(Y)???2??XY?15.设二维随机变量(X, Y )有联合概率密度函数
?1?(x?y),0?x?2, 0?y?2 f(x,y)??8?其它?0,试求E(X),E(Y),cov(X, Y),?XY. 解:E(X)???????????xf(x,y)dxdy?1227x(x?y)dxdy?, 8?0?06 2
概率论与数理统计统计课后习题答案-总主编-邹庭荣-主编-程述汉-舒兴明-第四章
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