定量法第一章习题解答 习题2 (1)P{恰有一件次品} 6194 0.0855 200 (2)P{全是正品} 194 0.9122 200 (3)P{至少2件正品}=P{2件正品}+P3件正品} P{1件次品}P{全是正品} =0.0855+0.9122=0.9978
定量法第一章习题解答 习题2. (1) P{恰有一件次品} = =0.0855 (2) P{全是正品}= =0.9122 (3) P{至少2件正品}= P{2件正品}+ P{3件正品} = P{1件次品}+ P{全是正品} =0.0855+0.9122=0.9978 3 200 2 194 1 6 C C C 3 200 3 194 C C
习题4解答 设A={寿命≥50},B={寿命≥70}, 由题意,P(A=1-0.1=0.9, P(B)=1-0.75=0.25 求P(BA) BCA,∴P(AB)P(B) P(BA=P(AB)/P(A)=PB)/P(A 0.25/0.9=0.278
习题4解答 设A={寿命50},B={寿命70}, 由题意,P(A)=1-0.1=0.9, P(B)=1-0.75=0.25 求P(B|A)。 BA,∴P(AB)=P(B) P(B|A)=P(AB) / P(A)=P(B) / P(A) =0.25/0.9=0.278
习题6解答 设A1、A2、A3分别为抽到用甲、乙、丙厂原料生 产的产品,B={抽到次品},则 P(A1)=0.6,P(A2)=0.3,P(A3)=0.1 P(B|A1)=0.02,P(BA2)=0.03,P(B|A3)=0.05 (1)P(BA3)=P(A3)P(BA3)=0.1×0.05=0.005 (2)P(B)=P(A1)P(BA1)+P(A2)P(BA2) +P(A3)P(B|A3) =0.6×0.02+0.3×003+0.1×0.05=0.026 (3)P(A1B)=P(A1)P(BA1)/P(B) =0.6×0.02/0.026=04615
习题6解答 设A1、A2、A3分别为抽到用甲、乙、丙厂原料生 产的产品,B={抽到次品},则 P(A1 )=0.6,P(A2 )=0.3,P(A3 )=0.1 P(B|A1 )=0.02,P(B|A2 )=0.03,P(B|A3 )=0.05 (1) P(BA3 )=P(A3 )P(B|A3 )=0.1×0.05=0.005 (2) P(B)=P(A1 )P(B|A1 )+ P(A2 )P(B|A2 ) + P(A3 )P(B|A3 ) =0.6×0.02+0.3×0.03+0.1×0.05=0.026 (3) P(A1 |B)=P(A1 )P(B|A1 ) / P(B) =0.6×0.02 / 0.026=0.4615
习题7解答 设至少应配备n门高炮,并设X为击中敌机的高炮 数,则Ⅹ~B(n,O.02),由题意,使 P(X≥1)=1-P(X=0 =1-C0.02098″=1-098″≥0.3 得0.98007 nlg0.98-g0.7 n≥ 0.7 =1765 lg0.98 故至少应配备18门高炮
习题7解答 设至少应配备n门高炮,并设X为击中敌机的高炮 数,则X~B(n, 0.02),由题意,使 P(X≥1)=1-P(X=0) 得 0.98n≤0.7 n lg0.98≤lg0.7 故至少应配备18门高炮。 17.65 lg 0.98 lg 0.7 n = 1 0.02 0.98 1 0.98 0.3 0 0 = − = − n n Cn
习题8解答 (1)设X={一合中的次品数},则X~B(n,p), n=100,p=0.01,q=0.99 P{X=0}=C10×0010×099100366 (2)P{X>2}=1-P{X=0}-P{X=1}-P{X=2} 1-0.366-C10X×001×0.99C70×0.012×0998 =0.0794
习题8解答 (1) 设X={一合中的次品数},则X~B(n,p), n=100,p=0.01,q=0.99 P{X=0}= ×0.010×0.99100=0.366 (2) P{X>2}=1-P{X=0}-P{X=1}-P{X=2} =1-0.366- ×0.01×0.9999 - ×0.012×0.9998 =0.0794 0 C100 1 C100 2 C100
(3)设每合至少装n个钻头,用泊松分布求解, =np=0.01×n≈1 由题意,每合次品数Xn-100}=1-P{X>n99} e ≥0.98,即:∑=,≤0.02 k=7-99k! k=n-99 查泊松分布表(=1),得 (n99)mi=4,即:n-99=4,n=103,即每合至少装103个。 P{X≤3}=P{X=0}+P{X=1}+P{X=2}+P{X=3} +C×0.010×0.991030103×001×099102 0 103 ×0.012×0.99101C; ×0.013×0.99100 103 =0.9798
(3) 设每合至少装n个钻头,用泊松分布求解, 则: λ=np=0.01×n≈1 由题意,每合次品数X≤n-100 P{X≤n-100}=1-P{X>n-100}=1-P{X≥n-99} =1- , 即: 查泊松分布表(λ=1),得 (n-99)min=4,即:n-99=4, n=103,即每合至少装103个。 P{X≤3}=P{X=0}+P{X=1}+P{X=2}+P{X=3} = ×0.010×0.99103+ ×0.011×0.99102 + ×0.012×0.99101+ ×0.013×0.99100 =0.9798 0.98 99 ! = − − k n k k e 0.02 99 ! = − − k n k k e 0 C103 1 C103 2 C103 3 C103
习题9解答 (1)P(X≤1000)=F(1000 1000 )=1-e=0632 (2)P(X≥2000)=1-F(2000) 2000 e1000 )=e2=0.135 (3)P(500<X≤1500)=F(1500-F(500) 1-e-1.5-(1-e0.5)=e0.5-e1.5=0.383
习题9解答 (1) P(X≤1000)=F(1000) = (2) P(X≥2000)=1-F(2000) =1- (3) P(500<X≤1500)=F(1500)-F(500) =1-e -1.5-(1-e -0.5)=e-0.5-e -1.5=0.383 (1 ) 0.135 2 2000 1000 1 − = = − − e e (1 ) 1 0.632 1 1000 1000 1 − = − = − − e e
习题山1解答 (1)P{X>1500}=1Φ( 1500-1000 500 =1-p(1)=1-0.8413=0.1587 (2)800+700=1500,∵‘{X≥1500}c{X800} P{X>1500,X800}=P{X>1500 由题意,即要求: P{X≥800,X≥1500)P{X≥1500) P{X21500X≥800 P{X≥800 P{X≥800} 其中:P{X>800}=1-Φ( 800-1000 500 1-(0.4)=Q(04)=06554 故P{X>1500X800}=0.1587/06554=02421
习题11解答 (1) P{X>1500}=1- =1-φ(1)=1-0.8413=0.1587 (2) 800+700=1500,∵{X≥1500}{X≥800}, ∴P{X≥1500,X≥800}=P{X≥1500} 由题意,即要求: P{X≥1500|X≥800}= 其中:P{X≥800}=1- =1-φ(-0.4)=φ(0.4)=0.6554 故P{X≥1500|X≥800}=0.1587 / 0.6554=0.2421 ) 500 1500 1000 ( − { 800} { 1500) { 800} { 800, 1500) = P X P X P X P X X ) 500 800 1000 ( −
(3)解法一: 设X1、Ⅹ2、X3分别为3个元件的寿命,则Ⅹ1、Ⅹ2、 X3相互独立,都服从N(1000,5002)。 P{至少有一个损坏}-1-P{全不损坏} 1-P{x121000,x21000,X3>1000} 1-P{X21000P{X2≥>1000P{X3≥1000} =1-(P{X≥1000})3=1-(1-9(0)3=1-0.53=0.875 解法二:设Y为1000小时内损坏的元件数, 由于各元件独立工作,故Y服从二项分布, n=3,p=P{<1000}=(0)=0.5,q=0.5 P{Y1}=1-PY=0}=1-C050×0.5-0.875
(3) 解法一: 设X1、X2、X3分别为3个元件的寿命,则X1、X2、 X3相互独立,都服从N(1000, 5002 )。 P{至少有一个损坏}=1-P{全不损坏} =1-P{X1≥1000,X2≥1000,X3≥1000} =1-P{X1≥1000}P{X2≥1000}P{X3≥1000} =1-(P{X≥1000})3=1-(1-φ(0))3=1-0.53=0.875 解法二:设Y为1000小时内损坏的元件数, 由于各元件独立工作,故Y服从二项分布, n=3,p=P{X<1000}=φ(0)=0.5,q=0.5 P{Y≥1}=1-P{Y=0}=1- ×0.50×0.53=0.875 0 C3
习题12解答 E(=EC X-E(X)、E(Y)-E(X) 0 D(X) D(X) D()=DO X-E(X)、D(X) D(X)(√DX)2
习题12解答 0 ( ) ( ) ( ) ) ( ) ( ) ( ) ( = − = − = D X E X E X D X X E X E Y E 1 ( ( )) ( ) ) ( ) ( ) ( ) ( 2 = = − = D X D X D X X E X D Y D