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a.) Construct a relative-frequency histogram for the chest circumference data, using classes based on a single value.

b.) The population mean and population standard deviation of the chest circumferences are 39.85 and 2.07, respectively. Identify the normal curve that should be used for the chest circumferences.

“The associated normal curve is the one whose parameters are the same as the mean and standard deviation of the variable…” (Weiss, pg. 256). Thus the parameters used to find the normal curve for chest circumferences are µ = 39.85 and σ = 2.07.

c.) Use the table on page 254 to find the percentage of militiamen in the survey with chest circumference between 36 and 41 inches, inclusive. Note: as the circumference were rounded to the nearest inch, you are actually finding the percentage of militiamen in the survey with chest circumference between 35.5 and 41.5 inches.

In order to find the percentage of militiamen with chest circumferences between 36 and 41 inches, we must add the frequency of these variables, and then divide by the total observations.

189+409+753+1062+1082+935=4430

4430/5732 = .7729

77.29%

d.) Use the normal curve you identified in part (b) to obtain an approximation to the percentage of militiamen in the survey with chest circumference between 35.5 and 41.5 inches. Compare your answer to the exact percentage found in part (c).

By utilizing the parameters for the normal curve, µ = 39.85 and σ = 2.07, we can obtain an approximation to the percentage of militiamen within the survey with chest circumferences between 35.5 and 41.5 inches.

First we will solve for standardized version of x or the standardized variable corresponding to x. (Weiss, pg. 258).

Z = ( x - µ ) / σ

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...written assignment draws on case study discussion exercises at the end of chapter. When preparing your assignment, please identify each answer clearly by question and its number. Case Study: Chest Sizes of Scottish Militiamen (p. 295): Answer a, b, c, d. You must calculate results by hand (though you may use any technology of your choice to verify your answers). a. Construct a relative-frequency histogram for the chest circumference data, using classes based on a single value. Size | Frequency | | | | | | | | | | | | | 33 | 3 | | | | | | | | | | | | 34 | 19 | | | | | | | | | | | | 35 | 81 | | | | | | | | | | | | 36 | 189 | | | | | | | | | | | | 37 | 409 | | | | | | | | | | | | 38 | 753 | | | | | | | | | | | | 39 | 1062 | | | | | | | | | | | | 40 | 1082 | | | | | | | | | | | | 41 | 935 | | | | | | | | | | | | 42 | 646 | | | | | | | | | | | | 43 | 313 | | | | | | | | | | | | 44 | 168 | | | | | | | | | | | | 45 | 50 | | | | | | | | | | | | 46 | 18 | | | | | | | | | | | | 47 | 3 | | | | | | | | | | | | 48 | 1 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | b. The population mean and population standard deviation of the chest circumferences are 39.85 and...

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...Chapter 6 Case Study: Chest size of Scottish Militiamen p. 295 a. Chest size (in) | Frequency | Relative Frequency | 33 | 3 | 0.000523 | 34 | 19 | 0.003315 | 35 | 81 | 0.014131 | 36 | 189 | 0.032973 | 37 | 409 | 0.071354 | 38 | 753 | 0.131368 | 39 | 1062 | 0.185276 | 40 | 1082 | 0.188765 | 41 | 935 | 0.163119 | 42 | 646 | 0.112701 | 43 | 313 | 0.054606 | 44 | 168 | 0.029309 | 45 | 50 | 0.008723 | 46 | 18 | 0.003140 | 47 | 3 | 0.000523 | 48 | 1 | 0.000174 | | 5732 | 1 | b. (μ=39.85, σ=2.07) The normal curve is the one whose parameters are the same as the mean and standard deviation variable, which are 39.85 and 2.07, respectively. Thus the required normal curve has parameters μ=39.85 and σ=2.07. c. Chest size (in) | Frequency | Relative Frequency | 33 | 3 | 0.000523 | 34 | 19 | 0.003315 | 35 | 81 | 0.014131 | 36 | 189 | 0.032973 | 37 | 409 | 0.071354 | 38 | 753 | 0.131368 | 39 | 1062 | 0.185276 | 40 | 1082 | 0.188765 | 41 | 935 | 0.163119 | 42 | 646 | 0.112701 | 43 | 313 | 0.054606 | 44 | 168 | 0.029309 | 45 | 50 | 0.008723 | 46 | 18 | 0.003140 | 47 | 3 | 0.000523 | 48 | 1 | 0.000174 | | 5732 | 1 | Utilizing the green highlighted section above, the following was calculated: 189+409+753+1062+1082+935 equals 4430. 4430 is divided by 5732 to equal 0.7728. The percentage of militiamen in the survey with chest circumference between 36 and 41 inches is 77.28%. d. z = x- μσ z = 35.5-39...

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...CALIFORNIA CALIFORNIA An Interpretive History TENTH EDITION James J. Rawls Instructor of History Diablo Valley College Walton Bean Late Professor of History University of California, Berkeley TM TM CALIFORNIA: AN INTERPRETIVE HISTORY, TENTH EDITION Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright © 2012 by The McGraw-Hill Companies, Inc. All rights reserved. Previous editions © 2008, 2003, and 1998. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1234567890 QFR/QFR 10987654321 ISBN: 978-0-07-340696-1 MHID: 0-07-340696-1 Vice President & Editor-in-Chief: Michael Ryan Vice President EDP/Central Publishing Services: Kimberly Meriwether David Publisher: Christopher Freitag Sponsoring Editor: Matthew Busbridge Executive Marketing Manager: Pamela S. Cooper Editorial Coordinator: Nikki Weissman Project Manager: Erin Melloy Design Coordinator: Margarite Reynolds Cover Designer: Carole Lawson Cover Image: Albert Bierstadt, American (born...

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