Chap05 - Solutions to End-of-Section and Chapter Review...

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Solutions to End-of-Section and Chapter Review Problems 5 CHAPTER 5 5.1 (a) Distribution A Distribution B X P(X) X*P(X) X P(X) X*P(X) 0 0.50 0.00 0 0.05 0.00 1 0.20 0.20 1 0.10 0.10 2 0.15 0.30 2 0.15 0.30 3 0.10 0.30 3 0.20 0.60 4 0.05 0.20 4 0.50 2.00 1.00 1.00 1.00 3.00 μ = 1.00 μ = 3.00 (b) Distribution A X ( X μ ) 2 P(X) ( X μ ) 2 * P(X) 0 (–1) 2 0.50 0.50 1 (0) 2 0.20 0.00 2 (1) 2 0.15 0.15 3 (2) 2 0.10 0.40 4 (3) 2 0.05 0.45 1.50 σ = ( X μ ) 2 P ( X ) = 1.22 (b) Distribution B X ( X μ ) 2 P(X) ( X μ ) 2 * P(X) 0 (–3) 2 0.05 0.45 1 (–2) 2 0.10 0.40 2 (–1) 2 0.15 0.15 3 (0) 2 0.20 0.00 4 (1) 2 0.50 0.50 1.50 σ = ( X μ ) 2 P ( X ) = 1.22 (c) Distribution A: Because the mean of 1 is greater than the median of 0, the distribution is right-skewed. Distribution B: Because the mean of 3 is less than the median of 4, the distribution is left- skewed. The means are different but the variances are the same. 5

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6 Chapter 5: Some Important Discrete Probability Distributions 5.2 (a) Distribution C Distribution D X P(X) X*P(X) X P(X) X*P(X) 0 0.20 0.00 0 0.10 0.00 1 0.20 0.20 1 0.20 0.20 2 0.20 0.40 2 0.40 0.80 3 0.20 0.60 3 0.20 0.60 4 0.20 0.80 4 0.10 0.40 1.00 2.00 μ = 2.00 1.00 2.00 μ = 2.00 (b) Distribution C X ( X μ ) 2 P(X) ( X μ ) 2 * P(X) 0 (–2) 2 0.20 0.80 1 (–1) 2 0.20 0.20 2 (0) 2 0.20 0.00 3 (1) 2 0.20 0.20 4 (2) 2 0.20 0.80 2.00 σ = ( X μ ) 2 P ( X ) = 2.00 = 1.414 Distribution D X ( X μ ) 2 P(X) ( X μ ) 2 * P(X) 0 (–2) 2 0.10 0.40 1 (–1) 2 0.20 0.20 2 (0) 2 0.40 0.00 3 (1) 2 0.20 0.20 4 (2) 2 0.10 0.40 1.20 σ = ( X μ ) 2 P ( X ) = 1.20 = 1.095 (c) Distribution C is uniform and symmetric; D is unimodal and symmetric. Means are the same but variances are different. 5.3 (a)-(c) X n (a) P(X) X*P(X) (X- μ ) 2 (X- μ ) 2 * P(X) 0 40 0.080 0.000 9.339136 0.74713088 1 100 0.200 0.200 4.227136 0.84542720 2 142 0.284 0.568 1.115136 0.31669862 3 66 0.132 0.396 0.003136 0.00041395 4 36 0.072 0.288 0.891136 0.06416179 5 30 0.060 0.300 3.779136 0.22674816 6 26 0.052 0.312 8.667136 0.45069107 7 20 0.040 0.280 15.555136 0.62220544 8 16 0.032 0.256 24.443136 0.78218035 9 14 0.028 0.252 35.331136 0.98927181 10 8 0.016 0.160 48.219136 0.77150618 11 2 0.004 0.044 63.107136 0.25242854 (b) Mean = 3.056 Variance = 6.06886400 (c) StDev = 2.46350644 6
Solutions to End-of-Section and Chapter Review Problems 7 5.3 (d) P ( X < 4) = P ( X = 0) + P ( X = 1) + P ( X = 2) + P ( X = 3) cont. = (0.080 + 0.200 + 0.284 + 0.132) = 0.696 (e) P ( X 4) = P ( X < 4) + P ( X = 4) = 0.696 + 0.072 = 0.768 (f) P ( X 4) = 1 – P ( X < 4) = 1 – 0.696 = 0.304 (g) P ( X = 4) = 0.072 (h) P ( X > 4) = 1 – P ( X 4) = 1 – 0.768 = 0.232 5.4 (a)-(b) X P(x) X*P(X) (X- μ X ) 2 (X- μ X ) 2 * P(X) 0 0.10 0.00 4 0.40 1 0.20 0.20 1 0.20 2 0.45 0.90 0 0.00 3 0.15 0.45 1 0.15 4 0.05 0.20 4 0.20 5 0.05 0.25 9 0.45 (a) Mean = 2.00 variance = 1.40 (b) Stdev = 1.18321596 5.5 (a)-(b) X P(x) X*P(X) (X- μ ) 2 (X- μ ) 2 * P(X) 0 0.32 0 1.6129 0.516128 1 0.35 0.35 0.0729 0.025515 2 0.18 0.36 0.5329 0.095922 3 0.08 0.24 2.9929 0.239432 4 0.04 0.16 7.4529 0.298116 5 0.02 0.1 13.9129 0.278258 6 0.01 0.06 22.3729 0.223729 (a) Mean = 1.27 variance = 1.6771 (b) Stdev = 1.29503 5.6 (a) X P(X) (b) X P(X) (c) X P(X) \$ – 1 21/36 \$ – 1 21/36 \$ – 1 30/36 \$ + 1 15/36 \$ + 1 15/36 \$ + 4 6/36 (d) \$ – 0.167 for each method of play 5.7 (a) E ( X ) = (0.4)(\$100) + (0.6)(\$200) = \$160 (b) E ( Y ) = (0.4)(\$200) + (0.6)(\$100) = \$140 (c) σ X = (0.4)(100 – 160) 2 + (0.6)(200 –160) 2 = 2400 = 48.99 (d) σ Y = (0.4)(200 –140) 2 + (0.6)(100 – 140) 2 = 2400 = 48.99 (e) σ XY = (0.4)(100 – 160)(200 – 140) + (0.6)(200 – 160)(100 – 140) = – 2400 (f) E ( X + Y ) = E ( X ) + E ( Y ) = \$160 + \$140 = \$300 (g) σ X + Y = 2400 + 2400 + 2(–2400) = 0 7

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8 Chapter 5: Some Important Discrete Probability Distributions 5.8 (a) E ( X ) = (0.2)(\$ – 100) + (0.4)(\$50) +(0.3)(\$ 200) + (0.1)(\$300) = \$90 (b) E ( Y ) = (0.2)(\$50) + (0.4)(\$30) + (0.3)(\$ 20) + (0.1)(\$20) = \$30 (c) 2 2 2 2 (0.2)( 100 90) (0.4)(50 90) (0.3)(200 90) (0.1)(300 90) 15900 126.10 X σ = - - + - + - + - = = (d) σ Y = (0.2)(50 – 30) 2 + (0.4)(30 – 30) 2 + (0.3)(20 – 30) 2 + (0.1)(20 – 30) 2 = 120 = 10.95 (e) σ XY = (0.2)( –100 – 90)(50 – 30) + (0.4)(50 – 90)(30 – 30) + (0.3)(200 – 90)(20 – 30) + (0.1)(300 – 90)(20 – 30) = –1300 (f) E ( X + Y ) = E ( X ) + E ( Y ) = \$90 + \$30 = \$120 (g) 15900 120 2( 1300) 13420 115.84 X Y σ + = + + - = = 5.9 (a) E ( P ) = (0.4)(\$50) + (0.6)(\$100) = \$80 (b) σ P = (.4) 2 (9000) + (.6) 2 (15000) + 2(.4)(.6)(7500) = 102.18 5.10 (a) E ( P ) = 0.3(105) + 0.7(35) = \$56 2 2 (0.3) (14,725) (0.7) (11,025) 2(0.3)(0.7)( 12,675) \$37.47 P σ = + + - = ( 29 ( 29 37.47 100% 66.91% 56 P CV E P σ = = = (b) E ( P ) = 0.7(105) + 0.3(35) = \$84 2 2 (0.7) (14,725) (0.3) (11,025) 2(0.7)(0.3)( 12,675) \$53.70 P σ = + + - = ( 29 ( 29 53.70 100% 63.93% 84 P CV E P σ = = = (c) Investing 50% in the Dow Jones index fund will yield the lowest risk per unit average return at ( 29 ( 29 10 100% 14.29% 70 P CV E P σ = = = .
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