概率论与数理统计课后习题答案刘舒强 金明爱 编

1、目录 第二章随机变量2 第四章数字特征4 1 第二章随机变量 7.设随机变量表示“废品数” ， 则 b(20,0.1)。已经发现2件废品说明废品数 2， ( 3| 2) = ( 3, 2) ( 2) = ( 3) ( 2) = 1 ( 3) 1 ( 1) = 1 ( = 0) ( = 1) = 0.23 () = (1+ 2+ 3+ 4) = 1 (1+ 2+ 3+ 4) = 1 (1* 2* 3* 4) = 方案二 设 =“出故障机器的台数” ， 则 (80,0.01)， 于是 () = ( 3) = 1 ( = 0) ( = 1) ( = 2) ( = 3) = 0.0175 19.当 0

2、时， () = ( ) = 0； 2 当0 1时， () = ( ) = () = 0 = 2/2; 当1 2时， () = ( ) = () = 1 0 + 1 (2 ) = 2 2 + 2 + 1 当2 时， () = ( ) = 1 所以 () = 0, 0 2/2,0 1 2 2 + 2 + 1,1 2 1, 2 20. (1)由lim 0+ () = lim 0+( + 2/2) = (0) lim + () =lim +( + 2/2) = 1 得 = 1, = 1， 所以 () = 0, 0 1 2/2, 0 (2) x的密度函数为 () = () = 0, = + () = 1

3、0 (500 100( ) 1 20 + 30 (500 + 300( ) 1 20 = 10 (600 100) 1 20 + 30 (200 + 300) 1 20 = 15 2 2+ 350 5250 所以由 9280有 15 2 2+ 350 5250 9280 化简得32 140 + 1612 0， 解得 20.6 26 (1)为保证期望利润不少于9280元， 最少进货量为21. (2) = () = 15 2 2+ 350 5250， () = 15 + 350 = 0 得 = 23.3,所以进货量为23时所获利润的期望值最大. 4 12. = + () = + 1 2 | = 0

4、, (被积函数为奇函数) 2= + 2() = 0 2 1 2 + + 0 2 1 2 = ( 2 1 2 )0 0 + ( 2 1 2 )+ 0 + + 0 = 0 + + 0 = ()0 + 0 + ()+ 0 + + 0 = 0 + + + 0 = 2 所以 = 2 ()2= 2 15.由题目可知 (4,)，表示在一次观察中大于 3的概率， 即 = ( 3 ) = 3 1 2 cos 2 = ( sin 2 ) 3 = 1 2 所以 (4,1/2)， 则 2 = + ( )2= 1 + 22= 5 17. = 0 (0.3 + 0.2) + 1 (0.1 + 0.1) + 2 (0 +

5、0.3) = 0.8 = 1 (0.3 + 0.1 + 0) + 2 (0.2 + 0.1 + 0.3) = 1.6 max, = max0,1 ( = 0, = 1) + max0,2 ( = 0, = 2) + max1,1 ( = 1, = 1) + max1,2 ( = 1, = 2) + max2,1 ( = 2, = 1) + max2,2 ( = 2, = 2) = 1 0.3 + 2 0.2 + 1 0.1 + 2 0.1 + 2 0 + 2 0.3 = 1.6 max, 2= max0,12 ( = 0, = 1) + max0,22 ( = 0, = 2) + max1,1

6、2 ( = 1, = 1) + max1,22 ( = 1, = 2) + max2,12 ( = 2, = 1) + max2,22 ( = 2, = 2) = 12 0.3 + 22 0.2 + 12 0.1 + 22 0.1 + 22 0 + 22 0.3 = 2.8 max, = max, 2 (max, )2= 2.8 2.56 = 0.24 18. = 1 0 (,) = 1 0 1 = 1 0 22 = 2 3 = 1 0 (,) = 1 0 1 = 1 0 1 2 2| = 0 ( ) = 1 0 (,) = 1 0 1 = 1 0 1 2 2| = 0 5 19.由题目可知(

7、, )的联合分布为 u v 12 14/90 24/91/9 ( = 1, = 1) = (max, = 1,min, = 1) = ( = 1, = 1) = ( = 1)( = 1) = 4/9; ( = 2, = 1) = (max, = 2,min, = 1) = ( = 2, = 1) + ( = 1, = 2) = ( = 2)( = 1) + ( = 1)( = 2) = 4/9 26. (1)因为( ) = + 2(, ) = 12， 所以(, ) = 1/2， = (, ) = 1/12 (2) (. ) = = 2.4， 所以 ( + ) = + + 2(, ) = 17.8 27. (1) ( + + ) = () + ( ) + () = 1 (2) (, ) = = 0 (,) = = 1 (,) = = 1/2 ( + + ) = () + ( ) + () + 2(, ) + 2(,) + 2(,) = 4 28.二维随机变量(, )的联合密度为 (,) = 2,0 1,1 1 0,其他 所