Solution Manual For Probability and Statistical Inference 8th Edition by Hall | All Chapters Included | Complete Latest Guide A+.

Solution Manual For Probability and Statistical Inference 8th Edition by Hall | All Chapters Included | Complete Latest Guide A+. Chapter 1 Probability 1.1 Basic Concepts 1.1-2 (a) S = {bbb, gbb, bgb, bbg, bgg, gbg, ggb, ggg}; (b) S = {female, male}; (c) S = {000, 001, 002, 003, . . . , 999}. 1.1-4 (a) Clutch size: 4 5 6 7 8 14 Frequency: 3 5 7 27 26 37 8 2 0 1 1 (b) x h(x) 0.05 0.10 0.15 0.20 0.25 0.30 2 4 6 8 10 12 14 Figure 1.1–4: Clutch sizes for the common gallinule (c) 9. 1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 Chapter 1 1.1-6 (a) No. Boxes: 4 5 6 7 8 Frequency: 7 9 5 2 4 4 2 2 1 1 (b) x h(x) 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 2 4 6 8 10 22 24 Figure 1.1–6: Number of boxes of cereal 1.1-8 (a) f (1) = 2 10, f (2) = 3 10, f (3) = 3 10, f (4) = 2 10. 1.1-10 This is an experiment. 1.1-12 (a) 50/204 = 0.245; 93/329 = 0.283; (b) 124/355 = 0.349; 21/58 = 0.362; (c) 174/559 = 0.311; 114/387 = 0.295; (d) Although James’ batting average is higher that Hrbek’s on both grass and artificial turf, Hrbek’s is higher over all. Note the different numbers of at bats on grass and artificial turf and how this affects the batting averages. 1.2 Properties of Probability 1.2-2 Sketch a figure and fill in the probabilities of each of the disjoint sets. Let A = {insure more than one car}, P (A) = 0.85. Let B = {insure a sports car}, P (B) = 0.23. Let C = {insure exactly one car}, P (C) = 0.15. It is also given that P (A ∩ B) = 0.17. Since P (A ∩ C) = 0, it follows that P (A ∩ B ∩ C 0 ) = 0.17. Thus P (A 0∩ B ∩ C 0 ) = 0.06 and P (A 0∩ B 0∩ C) = 0.09. 1.2-4 (a) S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, TTHH, HTHT, THTH, THHT, HTTT, THTT, TTHT, TTTH, TTTT}; (b) (i) 5/16, (ii) 0, (iii) 11/16, (iv) 4/16, (v) 4/16, (vi) 9/16, (vii) 4/16. 1.2-6 (a) 1/6; (b) P (B) = 1 − P (B 0 ) = 1 − P (A) = 5/6; (c) P (A ∪ B) = P (S) = 1. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Probability 3 1.2-8 (a) P (A ∪ B) = 0.4 + 0.5 − 0.3 = 0.6; (b) A = (A ∩ B 0 ) ∪ (A ∩ B) P (A) = P (A ∩ B 0 ) + P (A ∩ B) 0.4 = P (A ∩ B 0 ) + 0.3 P (A ∩ B) = 0.1; (c) P (A 0∪ B 0 ) = P [(A ∩ B) 0 ] = 1 − P (A ∩ B) = 1 − 0.3 = 0.7. 1.2-10 Let A ={lab work done}, B ={referral to a specialist}, P (A) = 0.41, P (B) = 0.53, P ([A ∪ B] 0 ) = 0.21. P (A ∪ B) = P (A) + P (B) − P (A ∩ B) 0.79 = 0.41 + 0.53 − P (A ∩ B) P (A ∩ B) = 0.41 + 0.53 − 0.79 = 0.15. 1.2-12 A ∪ B ∪ C = A ∪ (B ∪ C) P (A ∪ B ∪ C) = P (A) + P (B ∪ C) − P [A ∩ (B ∪ C)] = P (A) + P (B) + P (C) − P (B ∩ C) − P [(A ∩ B) ∪ (A ∩ C)] = P (A) + P (B) + P (C) − P (B ∩ C) − P (A ∩ B) − P (A ∩ C) + P (A ∩ B ∩ C). 1.2-14 (a) 1/3; (b) 2/3; (c) 0; (d) 1/2. 1.2-16 (a) S = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)}; (b) (i) 1/10; (ii) 5/10. 1.2-18 P (A) = 2[r − r( √ 3/2)] 2r = 1 − √ 3 2 . 1.2-20 Note that the respective probabilities are p 0, p1 = p 0/4, p2 = p 0/4 2 , . . .. X∞ k=0 p0 4 k = 1 p0 1 − 1/4 = 1 p0 = 3 4 1 − p0 − p1 = 1 − 15 16 = 1 16. 1.3 Methods of Enumeration 1.3-2 (4)(3)(2) = 24. 1.3-4 (a) (4)(5)(2) = 40; (b) (2)(2)(2) = 8. 1.3-6 (a) 4 µ 6 3 ¶ = 80; (b) 4(2 6 ) = 256; (c) (4 − 1 + 3)! (4 − 1)!3! = 20. 1.3-8 9P4 = 9! 5! = 3024. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 4 Chapter 1 1.3-10 S ={ HHH, HHCH, HCHH, CHHH, HHCCH, HCHCH, CHHCH, HCCHH, CHCHH, CCHHH, CCC, CCHC, CHCC, HCCC, CCHHC, CHCHC, HCCHC, CHHCC, HCHCC, HHCCC } so there are 20 possibilities. 1.3-12 3 · 3 · 212 = 36, 864. 1.3-14 µ n − 1 r ¶ + µ n − 1 r − 1 ¶ = (n − 1)! r!(n − 1 − r)! + (n − 1)! (r − 1)!(n − r)! = (n − r)(n − 1)! + r(n − 1)! r!(n − r)! = n! r!(n − r)! = µ n r ¶ . 1.3-16 0 = (1 − 1) n = Xn r=0 µ n r ¶ (−1) r (1)n−r = Xn r=0 (−1) r µ n r ¶ . 2 n = (1 + 1) n = Xn r=0 µ n r ¶ (1)r (1)n−r = Xn r=0 µ n r ¶ . 1.3-18 µ n n1, n2, . . . , ns ¶ = µ n n1 ¶µ n − n 1 n2 ¶µ n − n 1 − n 2 n3 ¶ · · · µ n − n 1 − · · · − ns−1 ns ¶ = n! n1!(n − n 1)! · (n − n 1)! n2!(n − n 1 − n 2)! · (n − n 1 − n 2)! n3!(n − n 1 − n 2 − n 3)! · · · (n − n 1 − n 2 − · · · − ns−1 )! ns!0! = n! n1!n2! . . . ns! . 1.3-20 (a) µ 19 3 ¶µ 52 − 19 6 ¶ µ 52 9 ¶ = 102, 486 351, 325= 0.2917; (b) µ 19 3 ¶µ 10 2 ¶µ 7 1 ¶µ 3 0 ¶µ 5 1 ¶µ 2 0 ¶µ 6 2 ¶ µ 52 9 ¶ = 7, 695 1, 236, 664= 0.00622. 1.3-22 µ 45 36 ¶ = 886,163,135.