Solutions Manual to accompany STATISTICS FOR ENGINEERS AND SCIENTISTS Third Edition by William Navidi Table of Contents Chapter 1 . 1 Chapter 2 . 12 Chapter 3 . 53 Chapter 4 . 73 Chapter 5 . 115 Chapter 6 . 133 Chapter 7 . 168 Chapter 8 . 188 Chapter 9 . 207 Chapter 10 . 227 1 SECTION 1.1 Chapter 1 Section 1.1 1. (a) The population consists of all the bolts in the shipment. It is tangible. (b) The population consists of all measurements that could be made on that resistor with that ohmmeter. It is conceptual. (c) The population consists of all residents of the town. It is tangible. (d) The population consists of all welds that could be made by that process. It is conceptual. (e) The population consists of all parts manufactured that day. It is tangible. 3. (a) False (b) True 5. (a) No. What is important is the population proportion of defectives; the sample proportion is only an approximation. The population proportion for the new process may in fact be greater or less than that of the old process. (b) No. The population proportion for the new process may be 0.12 or more, even though the sample proportion was only 0.11. (c) Finding 2 defective circuits in the sample. 7. A good knowledge of the process that generated the data. 9. (a) A controlled experiment (b) It is well-justi?ed, because it is based on a controlled experiment rather than an observational study. 2 CHAPTER 1 Section 1.2 1. False 3. No. In the sample 1, 2, 4 the mean is 7/3, which does not appear at all. 5. The sample size can be any odd number. 7. Yes. If all the numbers in the list are the same, the standard deviation will equal 0. 9. The mean and standard deviation both increase by 5%. 11. The total height of the 20 men is 20 × 178 = 3560. The total height of the 30 women is 30 × 164 = 4920. The total height of all 50 people is 3560 + 4920 = 8480. There are 20 + 30 = 50 people in total. Therefore the mean height for both groups put together is 8480/50 = 169.6 cm. 13. (a) All would be divided by 2.54. (b) Not exactly the same, because the measurements would be a little different the second time. 15. (a) The sample size is n = 16. The tertiles have cutpoints (1/3)(17) = 5.67 and (2/3)(17) = 11.33. The ?rst tertile is therefore the average of the sample values in positions 5 and 6, which is (44 + 46)/2 = 45. The second tertile is the average of the sample values in positions 11 and 12, which is (76 + 79)/2 = 77.5. (b) The sample size is n = 16. The quintiles have cutpoints (i/5)(17) for i = 1, 2, 3, 4. The quintiles are therefore the averages of the sample values in positions 3 and 4, in positions 6 and 7, in positions 10 and 11, and in positions 13 and 14. The quintiles are therefore (23 + 41)/2 = 32, (46 + 49)/2 = 47.5, (74 + 76)/2 = 75, and (82 + 89)/2 = 85.5. 3 SECTION 1.3 Section 1.3 Stem 0 1 1. (a) 2 3 4 5 6 7 8 9 10 11 12 13 Leaf 011112235677 235579 468 11257 14699 5 16 9 0099 0 7 7 0.45 0.4 (b) Here is one histogram. Other choices for the endpoints are possible. Relative Frequency 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0 (c) 0 (d) 2 4 6 8 Rainfall (inches) 10 2 4 12 6 8 10 Rainfall (inches) 14 Rainfall (inches) 15 10 The boxplot shows one outlier.