Homework02solutions

Homework02solutions - Homework 02 Solutions Problem 1....

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Homework 02 Solutions Problem 1. 2-26. a) Sample mean: 83.11, sample standard deviation = 7.11 b) Q 1 = 79.5, Q 3 = 84.50 d) Sample mean = 81, sample standard deviation = 3.46, Q 1 = 79.25, Q 3 = 83.75 Problem 2 2-32. a) 40 30 20 10 49 48 47 46 45 44 43 Index Vi scos it y 100 90 80 count
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Stem-and-leaf display for Problem 2-32.Viscosity: unit = 0.1 1|2 represents 1.2 2 42o|89 12 43*|0000112223 16 43o|5566 16 44*| 16 44o| 16 45*| 16 45o| 16 46*| 16 46o| 17 47*|2 (4) 47o|5999 19 48*|000001113334 7 48o|5666689 b) The plots indicates that the process is not stable and not capable of meeting the specifications. Problem 3 2-38. a) Both sample correlations will be negative. y versus x 1 40 30 20 10 0 200 150 100 x1 y
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y versus x 2 1400 1300 1200 1100 1000 900 800 200 150 100 x2 y b) y versus x 1 : -0.852; y versus x 2 : -0.898 Problem 4 3-11. a) P(X > 15) = 1 – P(X 15) = 1 – 0.3 = 0.7 b) P(X 24) = P(X 15) + P(15 < X 24) = 0.3 + 0.6 = 0.9 c) P(15 < X 20) = P(X 20) – P(X 15) = 0.5 – 0.3 = 0.2 d) P(X 18) = P(15 < X 18) + P(X 15) where P(15 < X 18) = P(15 < X 24) – P(18 < X 24) = 0.6 – 0.4 = 0.2 Therefore, P(X 18) = P(15 < X 18) + P(X 15) = 0.2 + 0.3 = 0.5 Alternatively, P(X 18) = P(X 24) - P(18 < X 24) = 0.9 – 0.4 = 0.5. Problem 5
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This note was uploaded on 10/11/2011 for the course CHEN 3010 taught by Professor Kompala,dh during the Spring '08 term at Colorado.

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Homework02solutions - Homework 02 Solutions Problem 1....

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