Probablity and Statistics Chp2 Solutions

Probablity and Statistics Chp2 Solutions - 2.21 Wch 11.1 =...

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Unformatted text preview: 2.21 Wch 11.1 = 6 sightseeing tours each available on 11.2 = 3 different days, the multiplication rule gives 11.1-11.2 : = 18 ways for a person to arrange a tour. 2.28 With 11.1 : 5 diiferent manufacturers, 11-2 : 3 different preparations, and 11-3 : 2 different strengths, the generalized multiplication rule yields 111112-103 : (5)(.‘3)(2) : 30 different ways to prescribe a drug for asthma. 2.33 Since the first digit is a 5, there are 11.1 = 9 possibilities for the second digit and then 11-; = 8 possibilities for the third digit. Therefore, by the multiplication rule there are 11.1-11.2 2 (9X8) 2 72 registrations to be checked. m; 2.53 S : {$10,$25,$100} with weights 275/500 : 11/20, 150/500 : 3/10, and 75/500 : 3 / 20, respectively. The probability that the first envelope purchased contains less than $100 is equal to 11/20 + 3/10 : 17/20. 2.56 Consider the events B: customer invests in tax—free bonds, 111’: customer invests in mutual funds. (a) P[B 051) = 13(8) + P151) — P(B r151) = 0.6 + 0.3 — 0.15 = 0.75. (b) P(B’ o 51') : 1 — P(B o .11) : 1 — 0.75 : 0.25. 2.58 (a) Let A = Defect in brake system; B = Defect in fuel system; PM U B) 2 PM) —|— 13(3) — MA o B) = 0.25 + 0.17 — 0.15 = 0.27. (b) P(N0 defect):1—P[AUB):1—0.27:0.73. 1V; *1”- 20? (a) 0.32; (b) 0.08; {c} office or den. 2.70 [3.) 0.02 + 0.30 = 0.32 = 32%; (b) 0.32 + 0.25 + 0.30 : 0.32 : 37%; (c) 0.05 + 0.00 + 0.02 : 0.13 : 13%; (d) 1 — 0.05 — 0.32 : 0.03 : 03%. 2.78 P[S | A) : 10/18 : 5/9. 2.79 Consider the et-‘ents: M: a, person is a, male; 3: EL person hes a, set-.ondstrj.I ednoetion; C: a person has a, college degree. (3,) P191 | is) = 29/19 = 14/39; (b) PH?” 91”) = 95/112. LEIJthLJlJ'LLIJ 4.134 J_l'JLI_-J.'L JIJL-IJ 444-. LLthlJII'LIJ. h 2.80 Consider the events: A: a person is experiencing hypertension, B: a person is a hast-’3I smoker, C: a person is a nonsnioker. (a) P(A | B) : 39/49; (b) P(C | A“) : 49/93 : 19/31. 2.84 Consider the events: C: an oil change is needed, F: an oil filter is needed. [a] P[F | o) = W“? : [L14 — use. ch'} ass — P on? . (bl Pic i F) : lam} : % Z 0'35- 2.88 Define H: head of household is home, C: a change is made in long distance carriers. P(H fl 0) = P(H)P(C | H) = (0.4)(03) = 0.12. 2.89 Consider the events: A: the doctor makes a correct diagnosis. B: the patient sues. P(A’ DB) 2 P[A’)P[B | A“) = [0.3)[0.9) = 02?. 2.101 Consider the events: C: an adult selected has cancer, D: the adult is diagnosed as having cancer. 13(0) : 0.05, P[D | C) : 0.78, PIC“) : 0.95 and P(D | C”) : 0.06. So, P[D) : P(C- H D) —|— P(C’ H D) : (0.05)(0.78) + (0.95)(0.06) : 0.096. 2.103 P(C | D) = = = 0.40625. 2.108 Consider the events: A: a customer purchases latex paint, A’ : a customer purchases semigloss paint, B: a. customer purchases rollers. _ PLB | .AthLAJ _ (Dfiflflflffi) _ H P(A I B) _ PEIB | -4)PEI-4jJ+P{B | A’jPLA’) — (0.60){0.75jn+(o.25){0.3o) — 0'85" ...
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This note was uploaded on 09/18/2011 for the course MAS 311 taught by Professor Hulett during the Spring '11 term at University of Miami.

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Probablity and Statistics Chp2 Solutions - 2.21 Wch 11.1 =...

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