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

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Unformatted text preview: 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) 0.50 0.00 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) (–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) (–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 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) 0.20 0.00 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) (–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) (–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) 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) 0.10 0.00 4 0.40 1 0.20 0.20 1 0.20 2 0.45 0.90 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)...
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This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue.

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

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