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: 9

Answer

a)0.9974 b)0.5086.

Work Step by Step

a) $z_{1}=\frac{value-mean}{standard \ deviation}=\frac{231.3-205.5}{8.6}=3$ $z_{2}=\frac{value-mean}{standard \ deviation}=\frac{179.7-205.5}{8.6}=-3$ Using the table, the probability of z being between -3 and 3 is equal to the probability of z being less than 3 minus the probability of z being less than -3, which is: 0.9987-0.0013=0.9974. b) By using the Central Limit Theorem, the sample mean has a mean of $\mu$ and standard deviation of $\frac{\sigma}{\sqrt n}$. $z_{1}=\frac{value-mean}{standard \ deviation}=\frac{206-205.5}{\frac{8.6}{\sqrt{40}}}=0.37.$ $z_{1}=\frac{value-mean}{standard \ deviation}=\frac{204-205.5}{\frac{8.6}{\sqrt{40}}}=-1.1.$ Using the table, the probability of z being between 0.37 and -1.1 is equal to the probability of z being less than 0.37 minus the probability of z being less than -1.1, which is: 0.6443-0.1357=0.5086.
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