《應用概率統(tǒng)計 課后答案》由會員分享，可在線閱讀，更多相關《應用概率統(tǒng)計 課后答案（47頁珍藏版）》請在裝配圖網(wǎng)上搜索。
1、課后答案網(wǎng)，用心為你服務！大學答案 - 中學答案 - 考研答案 - 考試答案最全最多的課后習題參考答案，盡在課后答案網(wǎng)（）！Khdaw團隊一直秉承用心為大家服務的宗旨，以關注學生的學習生活為出發(fā)點，旨在為廣大學生朋友的自主學習提供一個分享和交流的平臺。愛校園（） 課后答案網(wǎng)（） 淘答案（）!#%$%&%(%)%*%+-,-.0/1#%1%&%(%2%3%+-,04-561.798:p?ABC;GFDDEEHIpk = P (= k) = qk1p;k = 1; 2; :FKJq= 1p.2.67(1), L = 2 M;ONP( = 2) = pq + qp = 2pq, PRQSfUWVTX
2、YOZUWXgAGVYa L = 3 M;GNP( = 2) = p2q + q2p PRQSfUWVTVXYGZbUWXgA7XVYcd;Le= k M;P (= k) = pk1q + qk 1p = pq(pk 2 + qk 2;k = 2; 3; 4; :(2), L = k M;GfKgihkjkl(FVmp);Gnk1 jJiNr 1 jklVm;GrPFopm 11pr 1(1 p)k1(r1) = Ckr11pr qk r ;pk = P ( = k) = pCkrk = r + 1; r + 2; :;FKJq= 1 p.3.67(1), L 1 = k M;GfKgiskA
3、5tu?vwkx;GFA4ouy k +z1; :; 10, rC 4P ( 1 = k) =10 k;k = 1; 2; :; 6:C 510L 3 = k M;OfKgiskA5tu?Nvuy 1; z2;:; k1 ;OF2ouyk +z1; :; 10, rC 21C 2kP ( 3 = k) =k10; k = 3; 4; :; 8:C 5105(2), |M10Gu?f1=S1gT= f5 jtKJicjN1g ;Gr15P ( 1 = 1) =C5i95 i :X105i=1STf 1 = 10g = f5 jt10gt;G|1P (1 = 10) = 105 :1L k = 2
4、; :; 9 M;if 1 = kg = f 1 kg f 1 k 1g;15P ( 1k)=XC5iki(10k)5i ;10i=115P ( 1k1)=XC5i(k1)i 10(k1)5 i :10i=1qP ( 1 = k) =15k)5i(k1)i (11k)5i:C5iki (10X10 i=14.67 2 f2; 3; :; 12g, FDE23456789101112P ( = k)1234565432136363636363636363636365.67Bn(m)Nj?NjM;Gm jMn;AmmnP ( = n) =XP (Bn(m)jAm )P (Am )m=1n11pmq
5、n m 1(11=m=1 Cnmw )mXn11pk+1qn k 11=k=0 Cnm1 (1w )k+1X1p=pp(1)1)n 1wwppp= p 1 (1)n 1 + (1)n 1(1)n ;wwwFKJk= m1; p = 1p.67KiA;G2jf1;?2; :g. B Kikj;Gnk1 jJi;Ghkj. Ji;C ikj;ink1 jK;Ji;ihkjKiJJiAkKhkjJi;MBk ihk jJM1Akf = kg = B + C = A1B1 A2B2 :Ak 1Bk+ A1 B1A2 B2:Ak1 Bk1Ak Bk :2ipk=P (= k) = P (B) + P (
6、C )=(0:6 0:4)k 10:4 + (0:6 0:4)k 10:6 0:6=(0:24)k 10:76 = qk 1p; k = 1; 2; :Qp?0:76 ABC7DE A;Gj2?f0; 1; 2; :g. STf = 0g KhcjJi;Grp0 = P (= 0) = 0:4:k1 M;f = kg = A1 B1A2 B2:Ak1 Bk1Ak Bk + A1 B1 A2B2:Ak Bk Ak+1qNpk = P ( = k) = (0:6 0:4)k 1 0:6 0:6 + (0:6 0:4)k 0:4=(0:24)k 10:456;k = 1; 2; :; ADqEBCD
7、E7.7(1) g (x) =1x+h F (t)dt I100hRx) = 0, 30; (x0)(x)()., ;2; (+1) = 1; (10 , L x y M;= xy ( 0), ZF (t)dt#(x)(y) =h ZxF (t)dty1x+hy+hZyF (t)dt# :=h Zy+F (t)dt1y+ +hy+h y +2)FKJ01; 2 1, wcuJiD3 0 h;G(x)(y) = F (y + + 1h) F (y + 2h) 0 L h 0 MgiQ8h; (x) 720 ,DJi(x)= F (x + h), FKJ0 1 ;GK8limF (x) = 1;x
8、!+1x limF (x) = 0:!Q8lim(x) = 1 0 M; Rx+1 y1 dF (y) +1, 7Rx+1 y1 dF (y) = +1, KiZ+1y dF (y) Z+1y2 F (y)dy;xx11R+11F (y)dy = +1, qNxy2+1+11x 1F (y)dylimxdF (y) =lim F (x) +limy2R1x!0+Zxyx!0+x!0+x1F (x)2=lim F (x) +limx1x!0+x!0+x2=lim F (x) +limF (x) = 0:x!0+x!0+4610.7( ; ) q A Jic;G( ; ) A p.d.f.f (x
9、; y) = (1; (x; y)A= (1; 0 y 2x x2; 0x20;F20;F(A)(A)FKJ(A) = R02(2x x2 )dx = (x21x3 )j02 =4. | Ap.d.f.33Z(0;F1 f (x; y)dy =3(2x x2); 0x2f(x) =46 IGL0 x2 M; ADE?3x1Z0F (x) = P ( x) =(2t t2)dt =(3x2x3);44QF (x)=f (x)=8 4(3x2x3); 0 0;x01 2(2xx ); 0x2:=Z1f (x; y)dy =4:dx(0;2FdF (x)3611.7ABC A“h;G”AB L 0
10、x h M;ADE?112F (x) = P ( 0;: 1;(h x)2x0;0 h75,P (k) = P (1k) = P (1k) = 1P ( 1k) = 0:25, | P ( 1k) =0:r 1 k = 0; 29; k = 0:71.515.7P(). Ki1Xpk = P (= k) =P (= kj= j)P (= j):j=ki j = j B(j; p), r P ( = kj = j) = Cjk pk (1p)j k ;G1kkj k jpk =XCj p (1p)rj=kj!1j!kj k j=Xp (1 p)rj=kk!(jk)!j!=k pk e1j k(1
11、 p)j kXk!j=k(jk)!k pk=ek!( p)k=ek!QP(p)19.67F(x) Alim F (x) = A = 1,x!1e(1p)p;k = 0; 1; 2; :Q A = 1,f (x) = F 0(x) =( 0;F2x; 0 x 1620.7(1),Z xF (x) = P ( x) =f (t)dt8 2 x28 2 x2;0 x1;0 0;x00;x011=1123=12+ 2xx; 1 2xx1; 1 2 2:(2) P ( 1:3) = 1 F (1:3) = 12 1:3 +1(1:3)2 + 1 = 0:245;2P (0:2 1:2) = F (1:2
12、) F (0:2) = 2 1:21(1:2)211(0:2)2 = 0:66:22621.67(1)111F1 (y) = P ( 1 y) = P= 1 FyydF1 (y)11011f1(y) = f=fdyyyy2y12; 0 1 1= (2; 0 y 1=3y2( 0;F0;Fyyy(2)(2jjP ( y 0F (y) F ( y); y 0F(y) = P ( 0( f (y); y 0f(y) =dF1 (y)=0;y0=0;y0=8 2y; 0 y 0;y02y; 0 y 1(3):F3(y) = P e y = P ( ln y) = 1 F ( ln y):f3(y) =
13、dF1 (y)=1f ( ln y)dyy(2 ln y;0F= (2 ln y=0; y0; y ln y 122.67(1) ADE;e1 y 0xex 0;+(R+1x1= (1f (x; y)dx =xe(1+y)2 dx;y00;2y00;Z1R0y 0(1+y) ;y 0; f (x; y) = f (x)f (y), 7(2)f (x)=+1f (x; y)dy =x1 8xydy;0x 1=(4(xx3 ); 0x 1Z(0;F0;FRf (y)=+1f (x; y)dx =0y 8xydx;0y 1=(4y3; 0y 0x+111(xk11tk21e x etdt;x 0=
14、(xk11e x; x 0=k1 )k2 )0;k1 )0;x0x0R0Q624.7(1)f (x)f (y)k1 ; 1), k2 ; 1), f (x; y) =6 f (x)f (y), 7( 0;F( 0;2xF=+1f (x; y)dy =R2x2 +xydy; 0x1=2x2 +;0x1033Z( 0;F( 0;F=+1f (x; y)dx =R1x2 +xydx; 0y2=10y2036 (2 + y);Z(2) L 0 y 2 M;f (x; y)f j (xjy)=f (y)8(3)(4)FKJ8x2+xy(6x2 +2xy30;F0;F 1)=P ( ; ) 2 D) = ZZD f (x; y)dxdy12xy=Z0dx Z1 x x2 +dy31x2(1 + x) +x4 (1 x)2 dx=Z06154165=Z0x3 +x2 +xdx =6327211=P1; 1;P222 j2P12112
應用概率統(tǒng)計 課后答案
