mont4e_sm_ch03_sec08

mont4e_sm_ch03_sec08 - 3-96. Let X denote a geometric...

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Unformatted text preview: 3-96. Let X denote a geometric random variable with parameter p. Let q = 1-p. ∞ ∞ ⎛1⎞ 1 E ( X ) = ∑ x (1 − p ) x −1 p = p ∑ xq x −1 = p ⎜ 2 ⎟ = x =1 x =1 ⎝p ⎠ p V ( X ) = ∑ ( x − 1 ) 2 (1 − p) x −1 p = ∑ px 2 − 2 x + p x =1 x =1 1 p ∞ ∞ ( 1 p )(1 − p) x −1 = p ∑ x 2 q x −1 − 2∑ xq x −1 + x =1 ∞ ∞ ∞ x =1 ∑q x =1 ∞ x −1 = p ∑ x 2 q x −1 − x =1 ∞ 2 p2 + 1 p2 = p ∑ x 2 q x −1 − x =1 1 p2 1 p2 1 p2 1 p2 d = p dq ⎡ q + 2q 2 + 3q 3 + ...⎤ − ⎣ ⎦ d = p dq ⎡ q(1 + 2q + 3q 2 + ...) ⎤ − ⎣ ⎦ d = p dq ⎡ (1−qq )2 ⎤ − p12 = 2 pq(1 − q) −3 + p(1 − q) −2 − ⎣ ⎦ [ 2(1 − p) + p − 1] = (1 − p) = q = p2 p2 p2 Section 3-8 3-97. X has a hypergeometric distribution N=100, n=4, K=20 20 80 1 3 100 4 ( )( ) = 20(82160) = 0.4191 a.) P ( X = 1) = ( ) 3921225 b.) ( )( ) = 4845(1) = 0.001236 c.) P ( X = 4) = ( ) 3921225 20 80 4 0 100 4 P ( X = 6) = 0 , the sample size is only 4 d.) E ( X ) K ⎛ 20 ⎞ = 4⎜ ⎟ = 0 .8 N ⎝ 100 ⎠ ⎛ 96 ⎞ ⎛ N −n⎞ V ( X ) = np (1 − p )⎜ ⎟ = 4(0.2)(0.8)⎜ ⎟ = 0.6206 ⎝ 99 ⎠ ⎝ N −1 ⎠ = np = n 3-18 3-98. ( )( ) = (4 ×16 ×15 ×14) / 6 = 0.4623 ( ) (20 ×19 ×18 ×17) / 24 ( )( ) = 1 b) P ( X = 4 ) = = 0.00021 ( ) (20 ×19 ×18 ×17) / 24 a) P ( X = 1) = 4 1 16 3 20 4 4 16 40 20 4 c) P( X ≤ 2) = P ( X = 0) + P ( X = 1) + P ( X = 2) ( )( ) + ( )( ) + ( )( ) = () () () 4 0 16 4 20 4 4 1 16 3 20 4 4 2 16 2 20 4 = ⎛ 16×15×14×13 4×16×15×14 6×16×15 ⎞ + + ⎜ ⎟ 24 6 2 ⎝ ⎠ ⎛ 20×19×18×17 ⎞ ⎜ ⎟ 24 ⎝ ⎠ = 0.9866 d) E(X) = 4(4/20) = 0.8 V(X) = 4(0.2)(0.8)(16/19) = 0.539 3-99. N=10, n=3 and K=4 0.5 0.4 0.3 P(x) 0.2 0.1 0.0 0 1 2 3 x 3-100. (a) f(x) = ⎜ ⎜ ⎛ 24 ⎞⎛12 ⎞ ⎟/ ⎟⎜ ⎟ ⎟⎜ ⎝ x ⎠⎝ 3 − x ⎠ ⎛ 36 ⎞ ⎜⎟ ⎜3 ⎟ ⎝⎠ (b) µ=E(X) = np= 3*24/36=2 V(X)= np(1-p)(N-n)/(N-1) =2*(1-24/36)(36-3)/(36-1)=0.629 (c) P(X≤2) =1-P(X=3) =0.717 3-19 3-101. Let X denote the number of men who carry the marker on the male chromosome for an increased risk for high blood pressure. N=800, K=240 n=10 a) n=10 ( )( ) = ( )( ) = 0.1201 P( X = 1) = () 240 560 1 9 800 10 240! 560! 1!239! 9!551! 800! 10!790! b) n=10 P ( X > 1) = 1 − P ( X ≤ 1) = 1 − [ P ( X = 0) + P ( X = 1)] P( X = 0) = ( )( ) = ( () 240 560 0 10 800 10 P ( X > 1) = 1 − P ( X ≤ 1) = 1 − [0.0276 + 0.1201] = 0.8523 240! 560! 0!240! 10!550! 800! 10!790! )( ) = 0.0276 3-102. Let X denote the number of cards in the sample that are defective. a) ( )( ) = P( X = 0) = () 20 0 120 20 140 20 P( X ≥ 1) = 1 − P( X = 0) 120! 20!100! 140! 20!120! = 0.0356 P( X ≥ 1) = 1 − 0.0356 = 0.9644 b) P ( X ≥ 1) = 1 − P( X = 0) P ( X = 0) = ( )( ) = () 5 0 135 20 140 20 135! 20!115! 140! 20!120! = 135!120! = 0.4571 115!140! P ( X ≥ 1) = 1 − 0.4571 = 0.5429 3-103. N=300 (a) K=243, n=3, P(X=1)=0.087 (b) P(X≥1)=0.9934 (c) K=26+13=39 P(X=1)=0.297 (d) K=300-18=282 P(X≥1)=0.9998 Let X denote the count of the numbers in the state's sample that match those in the player's sample. Then, X has a hypergeometric distribution with N = 40, n = 6, and K = 6. 3-104. a) ( )( ) = ⎛ 40! ⎞ = 2.61× 10 ( ) ⎜ 6!34! ⎟ ⎝ ⎠ ( )( ) = 6 × 34 = 5.31× 10 b) P ( X = 5) = () () ( )( ) = 0.00219 c) P ( X = 4) = () P( X = 6) = 6 6 6 5 6 4 34 0 40 6 34 1 40 6 34 2 40 6 −1 −7 −5 40 6 d) Let Y denote the number of weeks needed to match all six numbers. Then, Y has a geometric distribution with p = 3-20 1 3,838,380 3-105. and E(Y) = 1/p = 3,838,380 weeks. This is more than 738 centuries! Let X denote the number of blades in the sample that are dull. a) P ( X ≥ 1) = 1 − P( X = 0) P ( X = 0) = ( )( ) = () 10 0 38 5 48 5 38! 5!33! 48! 5!43! = 38!43! = 0.2931 48!33! P ( X ≥ 1) = 1 − P( X = 0) = 0.7069 b) Let Y denote the number of days needed to replace the assembly. ( )( ) = c) On the first day, P ( X = 0) = () ( )( ) = On the second day, P ( X = 0) = () 2 0 46 5 48 5 6 0 42 5 48 5 P(Y = 3) = 0.29312 (0.7069) = 0.0607 46! 5!41! 48! 5!43! = 46!43! = 0.8005 48!41! = 42!43! = 0.4968 48!37! 42! 5!37! 48! 5!43! On the third day, P(X = 0) = 0.2931 from part a. Therefore, P(Y = 3) = 0.8005(0.4968)(1-0.2931) = 0.2811. 3-106. a) For Exercise 3-97, the finite population correction is 96/99. For Exercise 3-98, the finite population correction is 16/19. Because the finite population correction for Exercise 3-97 is closer to one, the binomial approximation to the distribution of X should be better in Exercise 3-97. b) Assuming X has a binomial distribution with n = 4 and p = 0.2, ( )0.2 0.8 = 0.4096 P( X = 4) = ( )0.2 0.8 = 0.0016 P( X = 1) = 4 1 1 3 4 4 4 0 The results from the binomial approximation are close to the probabilities obtained in Exercise 3-97. c) Assume X has a binomial distribution with n = 4 and p = 0.2. Consequently, P(X = 1) and P(X = 4) are the same as computed in part b. of this exercise. This binomial approximation is not as close to the true answer as the results obtained in part b. of this exercise. d) From Exercise 3-102, X is approximately binomial with n = 20 and p = 20/140 = 1/7. P( X ≥ 1) = 1 − P( X = 0) = ( )( ) ( ) 20 0 1 0 6 20 7 7 = 1 − 0.0458 = 0.9542 = 1 − 0.4832 = 0.5168 finite population correction is 120/139=0.8633 From Exercise 3-92, X is approximately binomial with n = 20 and p = 5/140 =1/28 20 1 P( X ≥ 1) = 1 − P( X = 0) = ( 0 )( 28 ) 0 27 20 28 () finite population correction is 120/139=0.8633 3-21 ...
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