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 315: 1

Answer

a) $0.9983$ b) $0.9370$ c) $0.8385$ d) $-0.52$ e) $0.1401$

Work Step by Step

a) We find the z-score: $z=\frac{2.93−0}{1}=2.93$ Then, using the table of z-scores, we can find that the corresponding probability is $0.9983$. b) We find the z-score: z=\frac{-1.53−0}{1}=−1.53 Then, using the table of z-scores, we can find that the corresponding probability is $1−0.0630=0.9370$ c) We find the z-scores: z=\frac{2.07−0}{1}=2.07 z=\frac{-1.07−0}{1}=−1.07 Then, using the table of z-scores, we can find that the corresponding probability is $0.9808−0.1423=0.8385.$ d) Using Microsoft Excel, we can see that this value is $-0.52$. e) $z=\sqrt{\frac{0.27−0}{1/\sqrt{16}}}=1.08$ Then, using the table of z-scores, we can find that the corresponding probability is $1−0.8599=0.1401.$
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