Essentials of Statistics (5th Edition)

Published by Pearson
ISBN 10: 0-32192-459-2
ISBN 13: 978-0-32192-459-9

Chapter 6 - Normal Probability Distributions - Review - Review Exercises - Page 309: 1

Answer

a) .9983 b) .9370 c) .8385 d) -.52 e) .1401

Work Step by Step

a) We find the z-score: $z = \frac{2.93-0}{1}=2.93$ Thus, using the table of z-scores, we find that this corresponds to a probability of $.9983$ b) We find the z-score: $z = \frac{-1.53-0}{1}=-1.53$ Thus, using the table of z-scores, we find that this corresponds to a probability of $1-.0630=.9370$ c) We find the z-scores: $z = \frac{2.07-0}{1}=2.07$ $z = \frac{-1.07-0}{1}=-1.07$ Thus, using the table of z-scores, we find that this corresponds to a probability of $.9808-.1423=.8385$ d) Using Microsoft Excel, we see that this value is -.52. e) $z = \frac{.27-0}{1/\sqrt{16}}=1.08$ Thus, using the table of z-scores, we find that this corresponds to a probability of $1-.8599=.1401$
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