Elementary Statistics (12th Edition)

Published by Pearson
ISBN 10: 0321836960
ISBN 13: 978-0-32183-696-0

Chapter 7 - Estimates and Sample Sizes - Review - Cumulative Review Exercises - Page 376: 9


a)13.35%, yes. b)160.195 and 189.805

Work Step by Step

a)$z=\frac{value-mean}{standard \ deviation}=\frac{185-175}{9}=1.11$ Using the table the probability is 1 minus the probability belonigng to 1.11, hence: P=1-0.8665=0.1335=13.35%. This is not that close to 0, hence a significant amount will be lost. b)By using the table, the z-score corresponding to 0.05 and 0.95: z=$\pm 1.645$. Hence the corresponding values:$mean+z⋅standard \ deviation=175-1.645⋅9=160.195.$ $mean+z⋅standard \ deviation=175+1.645⋅9=189.805.$
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