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Statistics Homework Solutions 88

# Statistics Homework Solutions 88 - 88 CHAPTER 4(b Let be...

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88 CHAPTER 4 (b) Let σ be the required standard deviation. The value of σ must be chosen so that the 1st percentile of the distribution is 65. The z -score of the 1st percentile is approximately z 2 33. Therefore 2 33 65 75 σ . Solving for σ yields σ 4 292 N/m 2 . (c) Let μ be the required value of the mean. This value must be chosen so that the 1st percentile of the distribution is 65. The z -score of the 1st percentile is approximately z 2 33. Therefore 2 33 65 μ 5. Solving for μ yields μ 76 65 N/m 2 . 19. Let a 1 σ and let b μ σ . Then Z aX b . Now b 1 σ μ μ σ 0, and a 2 σ 2 1 σ 2 σ 2 1. Equation (4.25) now shows that Z N 0 1 . 21. Let X be the lifetime of bulb A and let Y be the lifetime of bulb B. Then X N 800 100 2 and Y N 900 150 2 . (a) Let D Y X . The event that bulb B lasts longer than bulb A is the event that D 0. μ D μ Y μ X 900 800 100. Since X and Y are independent, σ D σ 2 X σ 2 Y 100 2 150 2 180 278. Since D is a linear combination of independent normal random variables, D is normally distributed. The z -score of 0 is 0
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