/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 The propagation of fatigue crack... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 7

The propagation of fatigue cracks in various aircraft parts has been the subject of extensive study in recent years. The accompanying data consists of propagation lives (flight hours/ \(10^{4}\) ) to reach a given crack size in fastener holes intended for use in military aircraft ("Statistical Crack Propagation in Fastener Holes Under Spectrum Loading, \(J\). Aircraft, 1983: 1028-1032): \(\begin{array}{rrrrrrrr}.736 & .863 & .865 & .913 & .915 & .937 & .983 & 1.007 \\\ 1.011 & 1.064 & 1.109 & 1.132 & 1.140 & 1.153 & 1.253 & 1.394\end{array}\) a. Compute and compare the values of the sample mean and median. b. By how much could the largest sample observation be decreased without affecting the value of the median?

Short Answer

Expert verified
Sample mean is 0.8047, median is 1.009. Largest observation can be decreased to 1.132 without affecting the median.

Step by step solution

01

Calculate the Sample Mean

To find the sample mean, sum all the observations and divide by the number of observations. The data values are: \(0.736, 0.863, 0.865, 0.913, 0.915, 0.937, 0.983, 1.007, 1.011, 1.064, 1.109, 1.132, 1.140, 1.153, 1.253, 1.394\). Add them up: \(12.875\). Divide by the number of data points, \(n = 16\), giving a sample mean of \(\frac{12.875}{16} = 0.8047\).
02

Calculate the Sample Median

To find the median, order the data set and find the middle value. With \(16\) observations, calculate the median as the average of the 8th and 9th ordered observations: \((1.007 + 1.011) / 2 = 1.009\).
03

Determine Effect of Decreasing the Largest Observation

Currently, the largest observation is \(1.394\), and the median is based on the 8th and 9th values, which are \(1.007\) and \(1.011\). Decreasing the largest value can't alter these middle values since they are unaffected by changes past the 8th and 9th positions. Thus, the largest observation can be decreased to any number above \(1.132\) (which is second largest) without changing the median.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Mean
The sample mean is a fundamental concept in probability and statistics. It's the arithmetic average of a set of values, and it provides a central value for the data set. To calculate the sample mean of our given data, we first sum up all the data points. In this problem, the values are:
  • 0.736, 0.863, 0.865, 0.913, 0.915, 0.937, 0.983, 1.007,
  • 1.011, 1.064, 1.109, 1.132, 1.140, 1.153, 1.253, 1.394
The total sum is 12.875, and there are 16 data points. Thus, the sample mean (\(\bar{x}\)) is computed as:\[\bar{x} = \frac{12.875}{16} = 0.8047\]This gives us a simple and quick measure of the central tendency of the crack propagation lives. The sample mean summarizes the data with a single number that represents the typical value of the set.
Sample Median
The sample median is another measure of central tendency in a data set. Unlike the mean, the median is the middle value that separates the higher half from the lower half of the data. For our data of crack propagation lives, the aim is to find the middle value. First, we arrange the data in ascending order. With 16 values, we need to average the 8th and 9th values to find the median. Those ordered values are:
  • 1.007 and 1.011
The sample median is calculated as:\[\text{Median} = \frac{1.007 + 1.011}{2} = 1.009\]This value indicates that half of the data lies below 1.009 and half above it. The median is a great measure of central tendency especially when there are outliers, as it isn't affected by extremely high or low values.
Data Analysis
Data analysis involves exploring and understanding data to summarize its main features. In our exercise, we analyze crack propagation lives by using the sample median and mean. Understanding these concepts helps in simplifying complex data and making informed decisions. One key point of analysis here is understanding the effect of extreme values on the median. The largest value in our data is 1.394. It's important to know that we can decrease this value significantly, even to just a bit above 1.132, without altering the sample median, as the median depends only on the middle values of the data. In statistical crack propagation analysis, such insights are crucial. They help in assessing the reliability and life expectancy of aircraft components without being misled by unusually high data points that might skew the average but leave the median unchanged. Engaging in data analysis allows one to derive deep insights from numerical information by using tools like the mean and median efficiently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The California State University (CSU) system consists of 23 campuses, from San Diego State in the south to Humboldt State near the Oregon border. A CSU administrator wishes to make an inference about the average distance between the hometowns of students and their campuses. Describe and discuss several different sampling methods that might be employed. Would this be an enumerative or an analytic study? Explain your reasoning.

The following categories for type of physical activity involved when an industrial accident occurred appeared in the article "Finding Occupational Accident Patterns in the Extractive Industry Using a Systematic Data Mining Approach" (Reliability Engr. and System Safety, 2012: 108-122): A. Working with handheld tools B. Movement C. Carrying by hand D. Handling of objects E. Operating a machine F. Other Construct a frequency distribution, including relative frequencies, and histogram for the accompanying data from 100 accidents (the percentages agree with those in the cited article): \(\begin{array}{llllllllllllll}\mathrm{A} & \mathrm{B} & \mathrm{D} & \mathrm{A} & \mathrm{A} & \mathrm{F} & \mathrm{C} & \mathrm{A} & \mathrm{C} & \mathrm{B} & \mathrm{E} & \mathrm{B} & \mathrm{A} & \mathrm{C} \\ \mathrm{F} & \mathrm{D} & \mathrm{B} & \mathrm{C} & \mathrm{D} & \mathrm{A} & \mathrm{A} & \mathrm{C} & \mathrm{B} & \mathrm{E} & \mathrm{B} & \mathrm{C} & \mathrm{E} & \mathrm{A} \\ \mathrm{B} & \mathrm{A} & \mathrm{A} & \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{C} & \mathrm{D} & \mathrm{F} & \mathrm{D} & \mathrm{B} & \mathrm{B} & \mathrm{A} & \mathrm{F} \\\ \mathrm{C} & \mathrm{B} & \mathrm{A} & \mathrm{C} & \mathrm{B} & \mathrm{E} & \mathrm{B} & \mathrm{D} & \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{E} & \mathrm{A} & \mathrm{A} \\ \mathrm{F} & \mathrm{C} & \mathrm{B} & \mathrm{D} & \mathrm{D} & \mathrm{D} & \mathrm{B} & \mathrm{D} & \mathrm{C} & \mathrm{A} & \mathrm{F} & \mathrm{A} & \mathrm{A} & \mathrm{B} \\\ \mathrm{D} & \mathrm{B} & \mathrm{A} & \mathrm{E} & \mathrm{D} & \mathrm{B} & \mathrm{C} & \mathrm{A} & \mathrm{F} & \mathrm{A} & \mathrm{C} & \mathrm{D} & \mathrm{D} & \mathrm{A} \\ \mathrm{A} & \mathrm{B} & \mathrm{A} & \mathrm{F} & \mathrm{D} & \mathrm{C} & \mathrm{A} & \mathrm{C} & \mathrm{B} & \mathrm{F} & \mathrm{D} & \mathrm{A} & \mathrm{E} & \mathrm{A} \\ \mathrm{C} & \mathrm{D} & & & & & & & & & & & & \end{array}\)

Exercise 34 presented the following data on endotoxin concentration in settled dust both for a sample of urban homes and for a sample of farm homes: \(\begin{array}{llrrrrrrrrrr}\mathrm{U}: & 6.0 & 5.0 & 11.0 & 33.0 & 4.0 & 5.0 & 80.0 & 18.0 & 35.0 & 17.0 & 23.0 \\ \mathrm{~F}: & 4.0 & 14.0 & 11.0 & 9.0 & 9.0 & 8.0 & 4.0 & 20.0 & 5.0 & 8.9 & 21.0 \\ & 9.2 & 3.0 & 2.0 & 0.3 & & & & & & & \end{array}\) a. Determine the value of the sample standard deviation for each sample, interpret these values, and then contrast variability in the two samples. [Hint: \(\Sigma x_{i}=237.0\) for the urban sample and \(=128.4\) for the farm sample, and \(\Sigma x_{i}^{2}=10,079\) for the urban sample and \(1617.94\) for the farm sample.] b. Compute the fourth spread for each sample and compare. Do the fourth spreads convey the same message about variability that the standard deviations do? Explain. c. The authors of the cited article also provided endotoxin concentrations in dust bag dust: U: \(34.049 .0 \quad 13.0 \quad 33.024 .024 .035 .0 \quad 104.034 .040 .0 \quad 38.0 \quad 1.0\) F: \(\quad 2.064 .0 \quad 6.0 \quad 17.0 \quad 35.011 .0 \quad 17.0 \quad 13.0 \quad 5.0 \quad 27.0 \quad 23.0\) \(28.0 \quad 10.0 \quad 13.0 \quad 0.2\) Construct a comparative boxplot (as did the cited paper) and compare and contrast the four samples.

a. Let \(a\) and \(b\) be constants and let \(y_{i}=a x_{i}+b\) for \(i=1,2, \ldots, n\). What are the relationships between \(\bar{x}\) and \(\bar{y}\) and between \(s_{x}^{2}\) and \(s_{y}^{2}\) ? b. A sample of temperatures for initiating a certain chemical reaction yielded a sample average \(\left({ }^{\circ} \mathrm{C}\right)\) of \(87.3\) and a sample standard deviation of \(1.04\). What are the sample average and standard deviation measured in "F? [Hint: \(F=\frac{2}{5} C+32\).]

Automated electron backscattered diffraction is now being used in the study of fracture phenomena. The following information on misorientation angle (degrees) was extracted from the article "Hbservations on the Faceted Initiation Site in the Dwell-Fatigue Tested Ti-6242 Alloy: Crystallographic Orientation and Size Effects" (Metallurgical and Materials Trans., 2006: 1507-1518). \(\begin{array}{lcccc}\text { Class: } & 0-<5 & 5-<10 & 10-<15 & 15-<20 \\\ \text { Rel freq: } & .177 & .166 & .175 & .136 \\ \text { Class: } & 20-<30 & 30-<40 & 40-<60 & 60-<90 \\ \text { Rel freq: } & .194 & .078 & .044 & .030\end{array}\) a. Is it true that more than \(50 \%\) of the sampled angles are smaller than \(15^{\circ}\), as asserted in the paper? b. What proportion of the sampled angles are at least \(30^{\circ} ?\) c. Roughly what proportion of angles are between \(10^{\circ}\) and \(25^{\circ} ?\) d. Construct a histogram and comment on any interesting features.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.