Introductory Statistics 9th Edition

Published by Wiley
ISBN 10: 1-11905-571-7
ISBN 13: 978-1-11905-571-6

Chapter 3 - Section 3.2 - Measures of Dispersion for Ungrouped Data - Exercises - Page 96: 3.37

Answer

a. range = 52 variance = 248.2605 standard deviation = 15.7563 b. coefficient of variation = 0.1189 * 100% = 11.89%

Work Step by Step

a. Range = Highest Value - Lowest Value = 159-107 = 52 Variance = $σ^2=(∑〖(x-μ)〗^2 )/n-1 =[∑(x^2) -(∑(x)^2/N)/n-1 =[356107-351390.1]/19 =248.2605 $ Standard Deviation = $\sqrt 248.2605 = 15.7563$ b. Coefficient of Variation=$15.7563 \div 132.55 \times 100% =11.89%$
This answer is currently locked

Someone from the community is currently working feverishly to complete this textbook answer. Don’t worry, it shouldn’t be long.