Elementary Statistics (12th Edition)

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

Chapter 6 - Normal Probability Distributions - Review - Cumulative Review Exercises: 3

Answer

a) 0.0630. b) 2643.24 c) 0.0005.

Work Step by Step

a)$z=\frac{value-mean}{standard \ deviation}=\frac{2500-3369}{567}=-1.53.$ Using the table, the probability of z being less than -1.53 is: 0.0630. b)By using the table, the z-score corresponding to 10%=0.1: z=-1.28. Hence the corresponding value:$mean+z⋅standard \ deviation=3369−1.28⋅567=2643.24.$ c)$z=\frac{value-mean}{standard \ deviation}=\frac{1500-3369}{567}=-3.3.$ Using the table, the probability of z being less than -3.3 is: 0.0005.
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