mChapter 5 Distribution – Exercise Answer
Q1 a) P(Z < 1.47) = 0.9292 b) P(Z > 1.74) = 1 – 0.9591 = 0.0409 c) P(1.47 < Z < 1.74) = 0.9591 – 0.9292 = 0.0299 d) P(Z < 1.47) + P(Z > 1.74) = 0.9292 + (1 – 0.9591) = 0.9701 Q2 a) P(– 1.47 < Z < 1.74) = 0.9591 – 0.0708 = 0.8883 b) P(Z < – 1.47) + P(Z > 1.74) = 0.0708 + 0.0409 = 0.1117 c) If P(Z > A) = 0.025, P(Z < A) = 0.975. A = + 1.96. d) 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.

Q3 a) P(Z > 1.18) = 1 – 0.8810 = 0.1190 b) P(Z < – 0.21) = 0.4129 c) P(– 1.96 < Z < – 0.21) = 0.4129 – 0.0250 = 0.3879 d) P(Z > A) = 0.1587, P(Z < A) = 0.8413. A = + 1.00 Q4 a) P(X > 44) = P(Z > – 1.5) = 1 – 0.0668 = 0.9332 b) P(X < 41) = P(Z < – 2.25) = 0.0122 c) P(X < A) = 0.05,

A = 50 – 1.645(4) = 43.42 d) P(Xlower < X < Xupper) = 0.60 P(Z < – 0.84) = 0.2005≈0.20 and P(Z < 0.84) = 0.7995 0.80

Xlower = 50 – 0.84(4) = 46.64 and Xupper = 50 + 0.84(4) = 53.36 Q5 a) P(X < 25) = P(Z < –2.016) = 0.0217 b) P(X > 50) = P(Z > 0.484) = 1-0.6844 = 0.3156 c) P(30 < X < 40) = P(–1.516 < Z < –0.516) = 0.3015 – 0.0643=0.2372 d) P(X < A) = 0.99

(For your information, between is the solution by using computer: ,
Chapter 5 Exercise Answer

A = $ 68.42)
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Q6 a) P(34 < X < 50) = P(– 1.33 < Z < 0) = 0.4082 b) P(X < 30) + P(X > 60) = P(Z < – 1.67) + P(Z > 0.83) = 0.0475 + (1.0 – 0.7967) = 0.2508 c) P(X > A) = 0.80 , P(Z < – 0.84) ≈0.20

A = 50 – 0.84(12) = 39.92 thousand miles or 39,920 miles d) The smaller standard deviation makes the Z-values larger (a) P(34 < X < 50) = P(– 1.60 < Z < 0) = 0.4452 (b) P(X < 30) + P(X > 60) = P(Z < – 2.00) + P(Z > 1.00) = 0.0228 + (1.0 – 0.8413) = 0.1815 (c) A = 50 – 0.84(10) = 41.6 thousand miles or 41,600 miles Q7 a) P(X 75,000) = P(Z > 0.2186) =1 – 0.5871 = 0.4129 c) P(X > 1,000,000) = P(Z > 1.6175) = 1 – 0.9474 = 0.0526 Q8 a)