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: 3


a)97.88% b)1742.57

Work Step by Step

a)$z=\frac{value-mean}{standard \ deviation}=\frac{1500-1634}{66}=-2.03.$ Using the table, the probability of z being more than -2.03 is equal to 1 minus the probability of z being less than -2.03, which is: 1-0.0212=0.9788=97.88%. b)By using the table, the z-score corresponding to 95%=0.95: z=1.645. Hence the corresponding value:$mean+z⋅standard \ deviation=1634+1.645⋅66=1742.57.$
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