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Solutions to End-of-Section and Chapter Review Problems 127CHAPTER 66.1(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 betweenA= – 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 = 0.18756.3(a)P(0 < Z< 1.08) = 0.8599 – 0.5 = 0.3599(b)P(Z< 0) + P(Z> 1.08) = 0.5 + (1.0 – 0.8599) = 0.6401(c)P(– 0.21 < Z< 0) = 0.5 – 0.4168 = 0.0832(d)P(Z< – 0.21) + P(Z> 0) = 0.4168 + 0.5 = 0.9168(e)P(Z< 1.08) = 0.8599(f)P(Z> – 0.21) = 1.0 – 0.4168 = 0.5832(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) = 0.55696.4(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 belowZ= 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
128 Chapter 6: The Normal Distribution and Other Continuous Distributions6.5(c) ZX85 – 10010= – 1.50cont.P(75 < X< 85) = P(– 2.50 < Z< – 1.50) = 0.0668 – 0.0062 = 0.0606(d)ZX112 – 10010= 1.20P(X> 112) = P(Z> 1.20) = 1 – P(Z< 1.20) = 1.0 – 0.8849 = 0.1151(e)ZX80 – 10010–2.00ZX110 – 100101.00P(X< 80) = P(Z< – 2.00) = 0.0228P(X> 110) = P(Z> 1.00) = 1 – P(Z< 1.00) = 1.0 – 0.8413 = 0.1587P(X< 80) + P(X> 110) = 0.0228 + 0.1587 = 0.1815(f)P(X< A) = 0.10, P(Z< – 1.28) = 0.10Z= –100–1.2810ASolving forA, A= 100 – 1.28(10) = 87.20(g)P(Xlower< X< Xupper) = 0.80P(– 1.28 < Z) = 0.10 andP(Z< 1.28) = 0.90Z–1.28Xlower–10010Z1.28Xupper– 10010Xlower= 100 – 1.28(10) = 87.20 and Xupper= 100 + 1.28(10) = 112.80(h)P(X> A) = 0.7, so P(X< A) = 0.3–100–0.5210AZA= 100 – 0.52(10) = 94.86.6
Solutions to End-of-Section and Chapter Review Problems 1296.7
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 trucks(e)P(X> A) = 0.80P(Z < – 0.84) 0.20– 50–0.8412AZA= 50 – 0.84(12) = 39.92 thousand miles or 39,920 miles(f)The larger standard deviation makes the Z-values smaller.(a)P(34 < X< 50) = P(– 1.60 < Z< 0) = 0.4452(b)P(34 < X< 38) = P(– 1.60 < Z< – 1.20) = 0.1151 – 0.0548= 0.0603(c)P(X< 30) + P(X> 60) = P(Z< – 2.00) + P(Z> 1.00)= 0.0228 + (1.0 – 0.8413) = 0.1815(d)1000(1 – 0.1815) = 818.5 trucks(e)A= 50 – 0.84(10) = 41.6 thousand miles or 41,600 miles6.9

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