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 - Beyond the Basics - Page 296: 23


a) 2.05, yes. b)0.5947.

Work Step by Step

a) The sample is without replacement and 50 is more than 5% of 275, hence the correct value is: $\sigma_{\overline{x}}=\frac{N-n}{N-1}\cdot\frac{\sigma}{\sqrt {n}}=\frac{275-50}{275-1}\cdot\frac{16}{\sqrt {50}}=2.05.$ b) By using the Central Limit Theorem, the sample mean has a mean of $\mu$ and standard deviation of $\frac{\sigma}{\sqrt n}$. $z_{1}=\frac{value-mean}{standard \ deviation}=\frac{105-95.5}{2.05}=4.62.$ $z_{2}=\frac{value-mean}{standard \ deviation}=\frac{95-95.5}{2.05}=-0.24.$ Using the table, the probability of z being between 4.62 and -0.24 is equal to the probability of z being less than 4.62 minus the probability of z being less than -0.24, which is: 0.9999-0.4052=0.5947.
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