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Compute the sample median, 25% trimmed mean, 10% trimmed

Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye ISBN: 9780321629111 32

Solution for problem 40E Chapter 1

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 40E

Compute the sample median, 25% trimmed mean, 10% trimmed mean, and sample mean for the lifetime data given in Exercise 27, and compare these measures. Reference exercise-27 The paper “Study on the Life Distribution of Microdrills” (J. of Engr. Manufacture, 2002: 301–305) reported the following observations, listed in increasing order, on drill lifetime (number of holes that a drill machines before it breaks) when holes were drilled in a certain brass alloy. a. ?Why can a frequency distribution not be based on the class intervals 0–50, 50–100, 100–150, and so on? b. ?Construct a frequency distribution and histogram of the data using class boundaries 0, 50, 100, . . . , and then comment on interesting characteristics. c?. ?Construct a frequency distribution and histogram of the natural logarithms of the lifetime observations, and comment on interesting characteristics. d?. ?What proportion of the lifetime observations in this sample are less than 100? What proportion of the observations are at least 200?

Step-by-Step Solution:
Step 1 of 3

Problem 40E Answer: Step1: We have The paper “Study on the Life Distribution of Micro Drills” (J. of Engr. Manufacture, 2002: 301–305) reported the following observations, listed in increasing order, on drill lifetime (number of holes that a drill machines before it breaks) when holes were drilled in a certain brass alloy. x 11 59 81 105 161 14 61 84 105 168 20 65 85 112 184 23 67 89 118 206 31 68 91 123 248 36 71 93 136 263 39 74 96 139 289 44 76 99 141 322 47 78 101 148 388 50 79 104 158 513 We need to compute the sample median, 25% trimmed mean, 10% trimmed mean, and sample mean for the lifetime data given in Exercise 27, and compare these measures. Step2: Consider the given data in Ascending Order x 11 59 81 105 161 14 61 84 105 168 20 65 85 112 184 23 67 89 118 206 31 68 91 123 248 36 71 93 136 263 39 74 96 139 289 44 76 99 141 322 47 78 101 148 388 50 79 104 158 513 Sample Median = 25th value + 26th value 91+93 2 = 2 = 184 2 = 92 Therefore,the sample median of the given data is 92. Step3: 1).Consider the given data in Ascending Order x 11 59 81 105 161 14 61 84 105 168 20 65 85 112 184 23 67 89 118 206 31 68 91 123 248 36 71 93 136 263 39 74 96 139 289 44 76 99 141 322 47 78 101 148 388 50 79 104 158 513 Trimmed mean percent = 25 100 = 0.25 Sample size n = 50 Now, Trimmed mean percent×sample size n = 0.25 ×50 = 12.5 13 Now we have to trim the given data by removing 13 lower observations and 13 upper observations.After removing 13+13 = 26 observations from the given data then we have i x 1 67 2 68 3 71 4 74 5 76 6 78 7 79 8 81 9 84 10 85 11 89 12 91 13 93 14 96 15 99 16 101 17 104 18 105 19 105 20 112 21 118 22 123 23 136 24 139 n n = 24 x i 2274 i=1 n xi i=1 25% trimmed mean is iven by x = n = 2274 24 = 94.75 Therefore,25% trimmed mean is 94.75. 2).Consider the given data in Ascending Order x 11 59 81 105 161 14 61 84 105 168 20 65 85 112 184 23 67 89 118 206 31 68 91 123 248 36 71 93 136 263 39 74 96 139 289 44 76 99 141 322 47 78 101 148 388 50 79 104 158 513 Trimmed mean percent = 100 = 0.10 Sample size n = 50 Now, Trimmed mean percent×sample size n = 0.10 ×50 = 5 Now we have to trim the given data by removing 5 lower observations and 5 upper observations.After removing 5+5 = 10 observations from the given data then we have i x 1 36 2 39 3 44 4 47 5 50 6 59 7 61 8 65 9 67 10 68 11 71 12 74 13 76 14 78 15 79 16 81 17 84 18 85 19 89 20 91 21 93 22 96 23 99 24 101 25 104 26 105 27 105 28 112 29 118 30 123 31 136 32 139 33 141 34 148 35 158 36 161 37 168 38 184 39 206 40 248 n n = 40 x i 4089 i=1 n i=1i 10% trimmed mean is iven by x = n 4089 = 40 = 102.225 Therefore,10% trimmed mean is 102.225. Step4: Consider the given data in Ascending Order x 11 59 81 105 161 14 61 84 105 168 20 65 85 112 184 23 67 89 118 206 31 68 91 123 248 36 71 93 136 263 39 74 96 139 289 44 76 99 141 322 47 78 101 148 388 50 79 104 158 513 n xi= 5963 i=1 x i=1i Sample mean is given byx = n = 5963 50 = 119.26 Therefore, the sample mean is 119.26. Conclusion: From the above calculation we have 25% of the trimmed mean is 94.75, 10% of the trimmed mean is 102.225 and the sample mean of the given data is 119.26.By comparing these three results sample mean of the given data is greater than both 25% of the trimmed mean and 10% of the trimmed mean.

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Chapter 1, Problem 40E is Solved
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Textbook: Probability and Statistics for Engineers and the Scientists
Edition: 9
Author: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye
ISBN: 9780321629111

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