M540_sol_HW5 - Chapter 3: Discrete Random Variables and...

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Chapter 3: Discrete Random Variables and Their Probability Distributions 35 Instructor’s Solutions Manual 3.33 a. . ) ( ) ( ) ( ) ( b a b Y aE b E aY E b aY E + = + = + = + µ b. . ] ) [( ] ) [( ] ) [( ) ( 2 2 2 2 2 2 σ a Y E a a aY E b a b aY E b aY V = = = + = + 3.34 The mean cost is E (10 Y ) = 10 E ( Y ) = 10[0(.1) + 1(.5) + 2(.4)] = $13. Since V ( Y ) .41, V (10 Y ) = 100 V ( Y ) = 100(.41) = 41. 3.35 With FS SS B = , P ( B ) = P ( SS ) + P ( FS ) = ( ) ( ) 4999 2000 5000 3000 4999 1999 5000 2000 + = 0.4 P ( B |first trial success) = 4999 1999 = 0.3999, which is not very different from the above. 3.36 a. The random variable Y does not have a binomial distribution. The days are not independent. b. This is not a binomial experiment. The number of trials is not fixed. 3.37 a. Not a binomial random variable. b. Not a binomial random variable. c. Binomial with n = 100, p = proportion of high school students who scored above 1026. d. Not a binomial random variable (not discrete). e. Not binomial, since the sample was not selected among all female HS grads. 3.38 Note that Y is binomial with n = 4, p = 1/3 = P (judge chooses formula B ). a. p ( y ) = () y y y 4 3 2 3 1 4 , y = 0, 1, 2, 3, 4. b. P ( Y 3) = p (3) + p (4) = 8/81 + 1/81 = 9/81 = 1/9. c. E ( Y ) = 4(1/3) = 4/3. d. V ( Y ) = 4(1/3)(2/3) = 8/9 3.39 Let Y = # of components failing in less than 1000 hours. Then, Y is binomial with n = 4 and p = .2. a. P ( Y = 2) = 2 2 ) 8 (. 2 . 2 4 = 0.1536. b. The system will operate if 0, 1, or 2 components fail in less than 1000 hours. So, P(system operates) = .4096 + .4096 + .1536 = .9728. 3.40 Let Y = # that recover from stomach disease. Then, Y is binomial with n = 20 and p = .8. To find these probabilities, Table 1 in Appendix III will be used. a. P ( Y 10) = 1 – P ( Y 9) = 1 – .001 = .999. b. P (14 Y 18) = P ( Y 18) – P ( Y 13) – .931 – .087 = .844 c. P ( Y 16) = .589. 3.41 Let Y = # of correct answers. Then, Y is binomial with n = 15 and p = .2. Using Table 1 in Appendix III, P ( Y 10) = 1 – P ( Y 9) = 1 – 1.000 = 0.000 (to three decimal places).
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36 Chapter 3: Discrete Random Variables and Their Probability Distributions Instructor’s Solutions Manual 3.42 a. If one answer can be eliminated on every problem, then, Y is binomial with n = 15 and p = .25. Then, P ( Y 10) = 1 – P ( Y 9) = 1 – 1.000 = 0.000 (to three decimal places). b. If two answers can be (correctly) eliminated on every problem, then, Y is binomial with n = 15 and p = 1/3. Then, P ( Y 10) = 1 – P ( Y 9) = 0.0085. 3.43 Let Y = # of qualifying subscribers. Then, Y is binomial with n = 5 and p = .7. a. P ( Y = 5) = .7 5 = .1681 b. P ( Y 4) = P ( Y = 4) + P ( Y = 5) = 5(.7 4 )(.3) + .7 5 = .3601 + .1681 = 0.5282. 3.44 Let Y = # of successful operations. Then Y is binomial with n = 5. a. With p = .8, P ( Y = 5) = .8 5 = 0.328. b. With p = .6, P ( Y = 4) = 5(.6 4 )(.4) = 0.259. c. With p = .3, P ( Y < 2) = P ( Y = 1) + P ( Y = 0) = 0.528. 3.45 Note that Y is binomial with n = 3 and p = .8. The alarm will function if Y = 1, 2, or 3. Thus, P ( Y 1) = 1 – P ( Y = 0) = 1 – .008 = 0.992. 3.46 When p = .5, the distribution is symmetric. When p < .5, the distribution is skewed to the left. When p > .5, the distribution is skewed to the right. 3.47 The graph is above. 0 5 10 15 20 0.00 0.05 0.10 0.15 y p(y) 3.48 a. Let Y = # of sets that detect the missile. Then, Y has a binomial distribution with n = 5 and p = .9. Then, P ( Y = 4) = 5(.9) 4 (.1) = 0.32805 and P ( Y 1) = 1 – P ( Y = 0) = 1 – 5(.9) 4 (.1) = 0.32805.
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M540_sol_HW5 - Chapter 3: Discrete Random Variables and...

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