Elementary Statistics (12th Edition)

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

Chapter 6 - Normal Probability Distributions - 6-3 Applications of Normal Distributions - Beyond the Basics - Page 271: 38


a) 2024.24 and 29.49 b)30.74

Work Step by Step

a) By using the table, the z-score corresponding to 0.95: z=0.52 and z= 1.645. Hence the corresponding values:$mean+z⋅standard \ deviation=1511+1.645⋅312=2024.24.$ $mean+z⋅standard \ deviation=21.1+1.645⋅5.1=29.49.$ b)$z=\frac{value-mean}{standard \ deviation}=\frac{2100-1511}{312}=1.89.$ Hence the corresponding value:$mean+z⋅standard \ deviation=21.1+1.89⋅5.1=30.74.$
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