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HW 9 Req HW Solutions - Chap 7

HW 9 Req HW Solutions - Chap 7 - Problem Solutions for...

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Problem Solutions for Required Homework #9 from Chapter 7, Section 7.5 Required homework: Problems 29 and 45 (problem 45 is from the Supplementary Exercises” on page 289). 29. Given: μ = $2.34 per gallon σ = $0.20 per gallon The sampling distribution of the sample mean: 1. E(xbar) = μ = $2.34 2. Standard error of the mean = σ /√n For n = 30, standard error of the mean = $0.20/√ 30 = $0.0365 For n = 50, standard error of the mean = $0.20/√ 50 = $0.0283 For n = 100, standard error of the mean = $0.20/√ 100 = $0.0150 For n = 50, standard error of the mean = $0.20/√ 100 = $0.0200 3. Shape is normal, by the CLT, because all the sample sizes are 30 a. With n = 30, want Prob ( μ – $0.03 ≤ xbar ≤ μ + $0.03) = Prob( $2.14 ≤ xbar ≤ $2.54) 82 . 0 0365 . 0 $ 03 . 0 $ - = = = x x z σ μ Prob ($2.14 x $2.54) = Prob (-0.82 z 0.82) = Prob ( z 0.82) – Prob ( z -0.82) = 0.7939 – 0.2061 = 0.5878 b. With n = 30, want Prob ( μ – $0.03 ≤ xbar ≤ μ + $0.03) = Prob( $2.14 ≤ xbar ≤ $2.54) 06 . 1 0283 . 0 $ 03 . 0 $ - = = = x x z σ μ ASW 5e Req HW #8 from Chap 7 1
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Prob ($2.14 x $2.54) = Prob (-1.06 z 1.06) = Prob ( z 1.06) – Prob ( z -1.06) = 0.8554 – 0.1446 = 0.7108 c. With n = 100, want Prob ( μ – $0.03 ≤ xbar ≤ μ + $0.03) = Prob( $2.14 ≤ xbar ≤ $2.54) 50 . 1 02 . 0 $ 03 . 0 $ - = = = x x z σ μ
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