《概率论与数理统计》课程电子教案（PPT课件讲稿）第六章 习题集

第 概率统

第六章习题课 《概率统计》

6+1 E(x)=∑E(X) 6+2 D(X)=E(X)-(EX)96+10+1 6+30+2 正确求解 D()=G=D)1 EX2-(EX n n 0+1(+1) 6+1 n+3(+2)2」m(0+3+2)

6 -1(3) 21 ( ) 1 ( ) 1 ++ =  = =  n i E Xi n E X 2 2 2 ) 21 ( 31 ( ) ( ) ( ) ++ − ++ = − =   D X E X E X ? 正确求解   2 2 2 ( ) ( ) 1 ( ) i i i EX EX n n D X n D X = = = −  2 2 2 ( 3)( 2) 1 ( 2) ( 1) 3 1 1 + + + =  ++ − ++ =      n n

仿P:191~-192求得E(S2)<2 a2与题无关,等于没求! 正确求解 6+1 E(S2)=0=D(X) (0+3)(+2)2 6-2(1) 9 CNP(1-p) k=0.1….N x怎能取两项概率?

2 2 2 ( 3)( 2) 1 ( ) ( ) + + + = = =     D Xi E S 仿P.191~192 求得 2 2 E(S ) = 2  与题无关，等于没求！ 正确求解 6-2 (1) ( )         = = −  = − k N x C p p x x k k N k i N n 0,1, , (1 ) , , , 1   ? xi 怎能取两项概率?

正确求解 x.=0.1.….N (1)Q i=1,2,…,nk是 再看 什么? ()F(x,…x)=cp1-P) (3)E(X)=∑B(X)=∑和=m E(S2)=和(1-p) 注意二项分布的参数是什么?

(3) np np n E X n E X n i n i =  i =  = =1 =1 1 ( ) 1 ( ) ( ) (1 ) 2 E S = np − p 正确求解 ( )       = =  = i n x N x x i n 1,2, , 0,1, , , , 1   (1)  2 1 0 1 1 1 0 ( , , )  (1 )  = − = − i x k N k N F x x C p p  i (2) 再看 k 是 什么？ 注意二项分布的参数是什么？

正确求解 (2)P(H1=x12…X1o=x0) ∏cwp3(1-p)x 10N-∑x10 p(1-p)∏ 个4 0,1,…,N,i=1,2,…,10 (3)E(X)=Np E(S2)=Np(1-p)

(2)   = − = −  −  = = − = = = = 1 0 1 1 0 1 0 1 1 1 1 0 1 0 1 0 1 1 0 1 (1 ) (1 ) ( , , ) i x N x N x i x x N x N i i i i i i i i p p C C p p P X x  X x x = 0,1,  ,N, i =1,2,  ,10. i (3) E(X ) = N p ( ) (1 ) 2 E S = N p − p 正确求解

3()9={(…x)x=1 x;岂能属于随机变量! 正确求解 ={(x2…,x)x∈R,=12,…,n (2)L(x,…,x)=I 2 2兀O (2) (3)E(X)=4,D(X)=,E(S2)=a

6-3 (1)  = (x1 ,  , xn ) xi R, i =1,2,  ,n 2 2 2 ( ) 1 1 2 1 ( , , )    − − = =  i x L x x e n i  n (2)  =  = − − n i i n x n e 1 2 2 2 1 2 ( ) (2 ) 1     (3) ( ) , ( ) , 2 n E X D X  =  = 2 2 E(S ) = (1)  = (x1 ,  , xn ) xi  X, i =1,2,  ,n i x 岂能属于随机变量！ 正确求解

6利用P200推论2,求得 10-1)(15-1) W 10+15-2 X-Y 0.3 P(X-|>03)=P( WV10115 P(7(23)032)=P(7(2)>7(23)=a T(23)=032=0424→a=0.6744 本题未必有 书后表 中查不 故不能用t分布 到此数E

6-6 P( X −Y  0.3) 利用P.200推论2， 求得 3 10 15 2 (10 1) 3 (15 1) 3 = + − −  + −  SW = ) 0.3 ( 1 5 1 1 0 1 1 5 1 1 0 1 +  + − = W W S S X Y P = ( (23)   (23)) = 2 P T T (23) 0.3 2 0.424 2 T = =  = 0.6744 本题未必有 故不能用 t 分布 3 2 2 2 S1 = S = 书后表 中查不 到此数 = P( T(23)  0.3 2 )

正确求解 由题设得X~N(20,0.3),Y~N(20,02) X-Y~N(0,0.5) P(Xx-y>03)=1-P(x-y503) =1-P(-0.3≤X-y≤0.3) =1-(0.3√2)+①(0.32) 2-2(0.3√2)=0.6744

正确求解 由题设得 X ~ N ( 20, 0.3), Y ~ N ( 20, 0.2) X −Y ~ N (0, 0.5) P( X −Y  0.3) =1− P( X −Y  0.3) =1− P(− 0.3  X −Y  0.3) =1−(0.3 2 ) + (−0.3 2) = 2 − 2(0.3 2 ) = 0.6744

Q(1-,,20 fx-(2)=F(2)=mle,20 E(X0)=()dz no n/ze n F(={F(),f1(2)=m10-2)e2,20 . E(X)=L x(a)dz=L n(1-eledz z nz(e-c 37c 0 (-1) n-2m-2。-(n-1)z +(-1)"e)d

6-7 ( ) 1 1 ( ) , ( ) 1 , 0 (1) = − − = −  − F z F z F x e x n x X  ( ) ( ) , 0 (1) (1) =  =  − f z F z n e z n z X X   ( )  ( ) , ( ) n X F z F z n = ( ) (1 ) , 0 1 ( ) = −  − − − f z n e e z z n z X n       n E X zf z dz n ze n z X 1 ( ) ( ) 0 0 (1) (1) = = =      − E X zf z dz n z e e dz z n z n  X n   −  − − = = − 0 0 1 ( ) ( ) ( ) (1 ) ( )      − − − − − = − + + 0 2 3 1 1 2 1 nz(e z Cn e z Cn e z  C e e dz n n z n n z n n ( 1) ( 1) ) 2 ( 1) 1 1 2 − − −  − −  − − + − + −

Q|= +-+-+… n k 这 两 或者 个处 E(Xom= m 2 案是 -ee zaz 相等 的 =n2∑Cn1(-1)ze+)ldz 0 k=0 n∑Cm(-y ∑Cm(D1 nk k=0 (k+1)2k

= = + + + + = n n k 1 k1 ) 1 31 21 (1 1   或者 E X n C e e zdz z z nk k n n    − −  −= − = − ( )   ( ) 0 10 ( ) 1 n C ze dz k z nk k k n ( 1 ) 0 10 1 ( 1 ) − +  −= −  = −   2 10 1 ( 1 ) 1 ( 1 ) + =  − −= − k n C nk k k n  k C n k k k n 1 ( 1 ) 1 1 = − = − 这两 个答 案是 相等的