# 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 μ σ = 70 – 100 10 = – 3.00 P ( X < 70) = P ( Z < – 3.00) = 0.00135 25
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 μ σ = 85 – 100 10 = – 1.50 P (75 < X < 85) = P (– 2.50 < Z < – 1.50) = 0.0668 – 0.0062 = 0.0606 (e) Z = X μ σ = 80 – 100 10 = –2.00 Z = X μ σ = 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 (h) P ( X > A ) = 0.7, so P ( X < A ) = 0.3 100 0.52 10 A Z - = - = A = 100 – 0.52(10) = 94.8 6.6 (a) P ( X > 43) = P ( Z > – 1.75) = 1 – 0.0401 = 0.9599 (b) P ( X < 42) = P ( Z < – 2.00) = 0.0228 (c) P ( X > 57.5) = P ( Z > 1.88) = 1.0 – 0.9699 = 0.0301 (d) P (42 < X < 48) = P (–2.00 < Z < – 0.50) = 0.3085 – 0.0228 = 0.2857 (e) P ( X < 40) = P ( Z < – 2.50) = 0.0062 P ( X > 55) = P ( Z > 1.25) = 1.0 – 0.8944 = 0.1056 P ( X < 40) + P ( X > 55) = 0.0062 + 0.1056 = 0.1118 (f) P ( X < A ) = 0.05, 50 1.645 4 A Z - = - = A = 50 – 1.645(4) = 43.42 (g) P ( X lower < X < X upper ) = 0.60 P ( Z < – 0.84) = 0.20 and P ( Z < 0.84) = 0.80 Z = –0.84 = X lower – 50 4 Z = + 0.84 = X upper – 50 4 X lower = 50 – 0.84(4) = 46.64 and X upper = 50 + 0.84(4) = 53.36 (h) P ( X > A ) = 0.85, so P ( X < A ) = 0.15 50 1.04 4 A Z - = - = A = 50 – 1.04(4) = 45.84 26
Solutions to End-of-Section and Chapter Review Problems 27 6.7 (a) P ( X < 25) = P ( Z < –1.116) = 0.1322 (b) P ( X > 50) = P ( Z > 1.384) = 0.0832 (c) P ( X > 75) = P ( Z > 3.884) = 0.0001 (d) P (30 < X < 40) = P (–0.616 < Z < 0.384) = 0.3806 (e) P ( X < A ) = 0.99 Z = 2.3263 = 36.16 10 A - A = 59.4234 (f) P ( X > A ) = 0.80 Z = –0.8416 = 36.16 10 A - A = 27.7438 (g) The normality assumption may be invalid if there are some households spending an abnormal amount on gourmet coffee, which will cause the distribution to be skewed to the right. Note : The above answers are obtained using PHStat. They may be slightly different when Table E.2 is used. 6.8 (a) P (34 < X < 50) = P (– 1.33 < Z < 0) = 0.4082 (b) P (34 < X < 38) = P (– 1.33 < Z < – 1.00) = 0.1587 – 0.0918 = 0.0669 (c) P ( X < 30) + P ( X > 60) = P ( Z < – 1.67) + P ( Z > 0.83) = 0.0475 + (1.0 – 0.7967) = 0.2508 (d) 1000(1 – 0.2508) = 749.2 2245 749 trucks (e) P ( X > A) = 0.80 P ( Z < – 0.84) 2245 0.20 50 0.84 12 A Z - = - = A = 50 – 0.84(12) = 39.92 thousand miles or 39,920 miles (f) The smaller standard deviation makes the Z -values larger.