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Unformatted text preview: ISOM 111 L6L7, Fall 2009 1 Homework 4 Solutions I . Problems 6.14 (p.244) of the textbook. Solution : (a) a(1) Because the population of all delivery times is normal, the population of all sample means is also normally distributed. a(2) The mean of all sample means μ ¯ X = μ = 45 minutes . a(3) The standard deviation of all sample means σ ¯ X = σ √ n = 6 √ 5 ≈ 2 . 68 . a(4) The interval is [ μ ¯ X ± 3 σ ¯ X ] = [45 ± 3 × 2 . 68] = [36 . 96 , 53 . 04] . (b) If the delivery process is operating effectively, then the delivery time will be smaller than 53.04 with a very high probability ( ≥ 99 . 73%). Now the sample mean delivery time is bigger than 53.04, so the delivery process is unlikely to be operating effectively. II . Problem 6.24 (p.247) of the textbook. Solution : (a) If the proportion of customer delight is p = 0 . 48, then the mean and standard deviation of all sample proportions with sample size n = 350 are μ ˆ p = p = 0 . 48 , and σ ˆ p = r p (1 p ) n...
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This note was uploaded on 01/01/2010 for the course ISOM ISOM111 taught by Professor Anthonychan during the Spring '09 term at HKUST.
 Spring '09
 AnthonyChan

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