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1、概率论与数理统计(刘建亚)习题解答第五章51解:(1)样本空间:Rn ;密度函数: f ( x , , x ) = (1) n exp-( xi -m)22s21n2ps(2)T1 , T3 , T4 是统计量(不含未知参数);T2 , T5 , T6 不是统计量(含未知参数)。52解: X P(l) , E ( X ) = l , D ( X ) = l E ( X ) = E D ( X ) = D (E ( S n2 ) = E= 1n= 1n531X i1E ( X i )= E ( X ) = ln=n1X) =1D ( X) =1D ( X ) = lnn 2niin1( X i
2、-) 2 = E 12 = 1XX i2 - XE ( X i2 ) - E ( X2 )nnn D ( X ) + E 2 ( X ) - D ( X ) + E 2 ( X ) nl + l 2 - ln + l 2 = n -n1ls2-m解: X N ( m,) ,X N (0,1)nsn(1)P- 10 2 = 1 - PX-102= 1 - PX -10X5()252525= 1 - P X2- 105 5 - P X2-105 -5= 1 - F (5) -F (-5) = 21 -F(5)= 2(1 - 0.9875) = 0.025(2)1P (max( X 1 , X 5
3、) 12) = 1 - P (max( X 1 , X5 ) 12)= 1 - P ( X 12)5 = 1 - F(12 -10)5 2= 1 - F (1)5 = 0.5785(3)P (min( X 1 , X 5 ) 8) = 1 - P (min( X 1 , X5 ) 8)= 1 - 1 - (1 - P ( X 8) 5 = 1 -F (- 1)5 = F (1)5 = 0.421554X k8X k2解: X k N (0, 0.22 N (0, 1)2= (2 X k2) ,c)= c(8)0.20.20.04k=12X k2aX k2aP ( X k a ) = P ( 2
4、450 -2500- 2500-1P 2450 = PX= PXX()250n250n250n5n) = 1 - 1 -F () = F () = 0.99= 1 -F (-nnn555nn2查表得:= 2.3325 2.33136556解: X E(l ),l = 0.0015-0.0015 xx 0 f ( x) =0.0015ex 002- 0.0015 xx 01 - eF ( x) = x 00 样本内的个体相互独立,则(1)没有失效的概率:66P ( A 800) =P ( X i 800) = 1 - P ( Xi 800)i =1i=1= 1 - F (800)6 = 0.00
5、07466(2)失效的概率:6P ( A 3000) = P ( X i 3000) = F (3000)6 = 1 - e- 0.00153000 6 = 0.935i=157解:略。58证明:略。5-9证明:略。5-10解: X n+1 与 X1 , , X n 相互独立,因此 X n+1 与 X 相互独立,已知s2) ,X n +1 N ( m , s 2 ), X N ( m,nn +1 s2 ) ,即X n +1 - X N (0, 1)则 X n +1 - X N (0,nn +1s2nnS 2而n c2 ( n -1),所以s23( X n +1 - X )n +n1 s2nS 2s2n( n -1)5-11证明:略。5-12解:略。5-13解:略。5-14解:略。5-15解:略。5-16解:略。=X n +1 - Xn -1 t ( n -1)n +1S n5-17解:略。4