Đề thi môn toán 2

Đề thi môn toán

CU 1 (2 im)a)LP Gii 1 mụn (1 im)Gii Toỏn 6 2Gii Anh vn 10 6Gii Toỏn&AV 4 p = C(34,3)/C(46,3) = 0,3942TNG S 46 34 (khụng gii mụn no)b) X ~ N( 5,2 1,96 )P(X >= 6 ) = 0,2839 (1 im)CU 2 (2 im)a) A, B xung khc. P(A) = 0,3 v P(B) = 0,4 (1 im)1/ ỳng, P(A|B) = P(AB)/P(B) = 0.2/ỳng, P(AB) = P(A) + P(B) = 0,7.3/Sai, P(AB) = 05/ỳng, P(A/B/) = P(AB)/ = P() = 14/ỳng, P(A/B/) = P(AB)/ = 1 - P(AB) = 1 - (P(A) + P(B)) = 0,3.b)KIN 1 KIN 2 KIN 3SP loi I 7 6 2SP loi II 3 4 8TNG S 10 10 10Cỏc bin c chn c kin K12, K13 v K23 l h y . (1 im)A l bin c 2 sn phm chn ra u l sn phm loi I.P(A) = P(A|K12)P(K12) + P(A|K13)P(K13) + P(A|K23)P(K23) = 0,2267 hay 17/75CU 3 (1 im)Xi (i = 1, 2) s sn phm loi I do mỏy th i sn xutX1 ~ B( 50 0,6 ) X2 ~ B( 50 0,7 )S tin thu c Y = 18(X1 + X2) + 14(100 - X1 - X2) = 4X1 + 4X2 + 1400VarY = 16VarX1 + 16VarX2 = 360 ngn ngCU 4 (2 im)Trng lng (gr) 150 250 350 450 550 650S trỏi 10 40 140 110 80 20 400(1 im)a) c lng trung bỡnh ca loi trỏi cõy nyTrung bỡnh mu = 417,50 grPhng sai mu = 12693,75Phng sai mu cú c = 12725,56 112,81 gr tin cy 1 - =95% z = 1,96Sai s =11,05à (406,45 428,55 ) gramb) Kim nh trung bỡnh 2 phớa.Trung bỡnh mu trỏi loi I = 507,14 grPhng sai mu trỏi loi I = 4353,74Phng sai mu cú c = 4374,57 66,14 grKim nh à =550 grMc ý ngha =0,03 z =2,17TCK z = -9,39|z| > zbỏc b H0Trng lng trung bỡnh ca trỏi loi I l 550gr khụng chp nhn c. (1 im)CU 5 (2 im)Kớch thc TT KT mu T s mu3000 400 350a) T l mu f = 0,8750 tin cy 1 - =97%Sai s =0,0359P(0,8391 0,9109) (2518 2733 ) sn phm A (1 im)b)TNG TH 21 23 27 29Ly mu khụng hon li, kớch thc 2, s cú C(4,2) = 6 mu, vi phõn phi phng sai mu c nh sau:S22,0 18,0 32,0 8,0 18,0 2,0P 1/6 1/6 1/6 1/6 1/6 1/6hay (1 im)S22,0 8,0 18,0 32,0P 1/3 1/6 1/3 1/6CU 6Bỏc b H0Bỏo cỏo ca nh mỏy khụng ỳng vi lụ hng trờn. (1 im)Xỏc sut 3 sinh viờn c chn khụng gii mụn no = =2220,051()27,3(3)7,8147kiiiinnpnp===>=uehforum.comPDFaid.Com#1 Pdf Solutions . (25 18 27 33 ) sn phm A (1 im)b)TNG TH 21 23 27 29 Ly mu khụng hon li, kớch thc 2, s cú C(4 ,2) = 6 mu, vi phõn phi phng sai mu c nh sau:S 22, 0 18,0 32, 0. = P(A|K 12) P(K 12) + P(A|K13)P(K13) + P(A|K23)P(K23) = 0 ,22 67 hay 17/75CU 3 (1 im)Xi (i = 1, 2) s sn phm loi I do mỏy th i sn xutX1 ~ B( 50 0,6 ) X2 ~ B(