Chap06 - Solutions to End-of-Section and Chapter Review...

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Solutions to End-of-Section and Chapter Review Problems 25 CHAPTER 6 6.1 (a) P ( Z < 1.57) = 0.9418 (b) P ( Z > 1.84) = 1 – 0.9671 = 0.0329 (c) P (1.57 < Z < 1.84) = 0.9671 – 0.9418 = 0.0253 (d) P ( Z < 1.57) + P ( Z > 1.84) = 0.9418 + (1 – 0.9671) = 0.9747 (e) P (– 1.57 < Z < 1.84) = 0.9671 – 0.0582 = 0.9089 (f) P ( Z < – 1.57) + P ( Z > 1.84) = 0.0582 + 0.0329 = 0.0911 (g) Since the Z -distribution is symmetric about its mean, half of the area will be above Z = 0. (h) If P ( Z > A ) = 0.025, P ( Z < A ) = 0.975. A = + 1.96. (i) If P (– A < Z < A ) = 0.6826, P ( Z < A ) = 0.8413. So 68.26% of the area is captured between A = – 1.00 and A = + 1.00. 6.2 (a) P ( Z > 1.34) = 1.0 – 0.9099 = 0.0901 (b) P ( Z < 1.17) = 0.8790 (c) P (0 < Z < 1.17) = 0.8790 – 0.5 = 0.3790 (d) P ( Z < – 1.17) = 0.1210 (e) P (– 1.17 < Z < 1.34) = 0.9099 – 0.1210 = 0.7889 (f) P (– 1.17 < Z < – 0.50) = 0.3085 – 0.1210 = 0.1875 6.3 (a) P ( Z < 1.08) = 0.8599 (b) P ( Z > – 0.21) = 1.0 – 0.4168 = 0.5832 (c) P (0 < Z < 1.08) = 0.8599 – 0.5 = 0.3599 (d) P ( Z < 0) + P ( Z > 1.08) = 0.5 + (1.0 – 0.8599) = 0.6401 (e) P (– 0.21 < Z < 0) = 0.5 – 0.4168 = 0.0832 (f) P ( Z < – 0.21) + P ( Z > 0) = 0.4168 + 0.5 = 0.9168 (g) P (– 0.21 < Z < + 1.08) = 0.8599 – 0.4168 = 0.4431 (h) P ( Z < – 0.21) + P ( Z > 1.08) = 0.4168 + (1 – 0.8599) = 0.5569 6.4 (a) P ( Z > 1.08) = 1 – 0.8599 = 0.1401 (b) P ( Z < – 0.21) = 0.4168 (c) P (– 1.96 < Z < – 0.21) = 0.4168 – 0.0250 = 0.3918 (d) P (– 1.96 < Z < 1.08) = 0.8599 – 0.0250 = 0.8349 (e) P (1.08 < Z < 1.96) = 0.9750 – 0.8599 = 0.1151 (f) Since the Z -distribution is symmetric about its mean, half of the area will be below Z = 0. (g) If P ( Z < A ) = 0.1587, A = – 1.00. (h) If P ( Z > A ) = 0.1587, P ( Z < A ) = 0.8413. So A = + 1.00. 6.5 (a) Z = X μ σ = 75 –100 10 = – 2.50 P ( X > 75) = P ( Z > – 2.50) = 1 – P ( Z < – 2.50) = 1 – 0.0062 = 0.9938 (b) Z = X = 70 – 100 10 = – 3.00 P ( X < 70) = P ( Z < – 3.00) = 0.00135 25
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26 Chapter 6: The Normal Distribution, Other Continuous Distributions, and Sampling Distributions 6.5 (c) Z = X μ σ = 112 – 100 10 = 1.20 cont. P ( X > 112) = P ( Z > 1.20) = 1 – P ( Z < 1.20) = 1.0 – 0.8849 = 0.1151 (d) Z = X = 85 – 100 10 = – 1.50 P (75 < X < 85) = P (– 2.50 < Z < – 1.50) = 0.0668 – 0.0062 = 0.0606 (e) Z = X = 80 – 100 10 = –2.00 Z = X = 110 – 100 10 = 1.00 P ( X < 80) = P ( Z < – 2.00) = 0.0228 P ( X > 110) = P ( Z > 1.00) = 1 – P ( Z < 1.00) = 1.0 – 0.8413 = 0.1587 P ( X < 80) + P ( X > 110) = 0.0228 + 0.1587 = 0.1815 (f) P ( X < A ) = 0.10, P ( Z < – 1.28) = 0.10 Z = 100 1.28 10 A - - = Solving for A , A = 100 – 1.28(10) = 87.20 (g) P ( X lower < X < X upper ) = 0.80 P ( Z < – 1.28) = 0.10 and P ( Z < 1.28) = 0.90 Z = –1.28 = X lower –100 10 Z = + 1.28 = X upper – 100 10 X lower = 100 – 1.28(10) = 87.20 and X upper = 100 + 1.28(10) = 112.80
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Chap06 - Solutions to End-of-Section and Chapter Review...

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