Elementary Statistics (12th Edition)

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

Chapter 6 - Normal Probability Distributions - Review - Review Exercises - Page 316: 2


a)7.93% b)1369.21

Work Step by Step

a)$z=\frac{value-mean}{standard \ deviation}=\frac{1605-1516}{63}=1.41.$ Using the table, the probability of z being more than 1.41 is equal to 1 minus the probability of z being less than 1.41, which is: 1-0.9207=0.0793=7.93%. b)By using the table, the z-score corresponding to 1%=0.01: z=-2.33. Hence the corresponding value:$mean+z⋅standard \ deviation=1516−2.33⋅63=1369.21.$
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