ps4_answer

ps4_answer - 10 = $2600 Since n > 30, the shape...

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1 Econ 41 Statistics for Economists Summer 2007 Chunming Yuan Problem Set 4 (Answers) Multiple Choice Summary Table Item Answer Item Answer 1. c 18. c 2. b 19. a 3. a 20. a 4. c 5. a 6. c 7. c 8. c 9. b 10. a 11. b 12. d 13. b 14. c 15. b 16. d 17. b Problem and Essay Solutions 1. a. P(1.07 z 3.02) = .4987 – .3577 = .1410 b. P(–2.22 z –.92) = .4868 – .3212 = .1656 c. P(–1.38 z 2.66) = .4162 + .4961 = .9123 2. For x = 580, z = (580 – 670) ÷ 120 = –.75 F o r x = 892, z = (892 – 670) ÷ 120 = 1.85 P(670 < x < 580) = P(–.75 < z < 1.85) = .2734 + .4678 = .7412 3. a. z = 1.46 c. z = –2.58 b. z = –2.66 d. z = 2.07 4. P( μ – 2 σ < < + 2 σ ) = P(–2 < z P( – 3 σ < < + 3 σ ) = P(–3 < z < 3) = .4987 + .4987 = .9974 Chebyshev’s theorem is correct in its prediction in both cases. 5. = μ = 124, Standard deviation of the sample mean = 20
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2 6. x = μ = $84,500, Standard deviation of the sample mean = $26,000
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Unformatted text preview: 10 = $2600 Since n &gt; 30, the shape of the sampling distribution of the sample mean is approximately normal because of the central limit theorem. 7. p =.20, Standard deviation of 500 8 2 ) )(. (. = p = .0179 a. For p = .17, z = (.17 .20) .0179 = 1.67 P( z &lt; 1.67) = .5000 .4525 = .0475 b. For p = .18, z = (.18 .20) .0179 = 1.12 For p = .23, z = (.23 .20) .0179 = 1.67 P(1.12 &lt; z &lt; 1.67) = .3686 + .4525 = .8211 c. For p = .18, z = (.18 .20) .0179 = 1.12 P( z &gt; 1.12) = .5000 + .3686) = .8686 8. In both cases, n 30, so the sampling distribution will be normal a. sample mean = 26, standard deviation = 30 12 = 2.1909 b. sample mean = 26, standard deviation = 200 12 = .8485...
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ps4_answer - 10 = $2600 Since n &amp;amp;gt; 30, the shape...

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