Elementary Statistics (12th Edition)

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

Chapter 6 - Normal Probability Distributions - 6-5 The Central Limit Theorem - Basic Skills and Concepts - Page 295: 18


a) $P_1=50.472 $ $P_{99}=104.528$ b) $0.999$ c) Part a

Work Step by Step

a) Using the table of positive z-values, we see that the 99th percentile corresponds to a value of $z=±2.33$. Hence: $P_1=77.5+(11.6)(−2.33)=50.472$ $P_{99}=77.5+(11.6)(2.33)=104.528$ b) We find the two z-scores: $z=\frac{70−77.5}{11.6/\sqrt{25}}=−3.23$ $z=\frac{85−77.5}{11.6/\sqrt{25}}=3.23$ Hence: $p=0.9994−0.0004=0.999 $ c) The results from part a) should be used, because this is just a cutoff for further testing. Hence, the parameters from part b) are too rare.
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