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Unformatted text preview: 396. Let X denote a geometric random variable with parameter p. Let q = 1p.
∞ ∞ ⎛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 38 397. 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 318 398. ( )( ) = (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) = 0.539 399. 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 3100. (a) f(x) = ⎜ ⎜ ⎛ 24 ⎞⎛12 ⎞ ⎟/ ⎟⎜ ⎟ ⎟⎜ ⎝ x ⎠⎝ 3 − x ⎠ ⎛ 36 ⎞ ⎜⎟ ⎜3 ⎟ ⎝⎠ (b) µ=E(X) = np= 3*24/36=2 V(X)= np(1p)(Nn)/(N1) =2*(124/36)(363)/(361)=0.629 (c) P(X≤2) =1P(X=3) =0.717 319 3101. 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 3102. 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
3103. 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=30018=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. 3104. 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 = 320 1 3,838,380
3105. 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)(10.2931) = 0.2811. 3106. a) For Exercise 397, the finite population correction is 96/99. For Exercise 398, the finite population correction is 16/19. Because the finite population correction for Exercise 397 is closer to one, the binomial approximation to the distribution of X should be better in Exercise 397. 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 397. 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 3102, 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 392, 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 321 ...
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