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: 10

Answer

a)0.2596 b)0.659

Work Step by Step

a) $z_{1}=\frac{value-mean}{standard \ deviation}=\frac{180-205.5}{8.6}=-2.97$ $z_{2}=\frac{value-mean}{standard \ deviation}=\frac{200-205.5}{8.6}=-0.64$ Using the table, the probability of z being between -2.94 and -0.64 is equal to the probability of z being less than -0.64 minus the probability of z being less than -2.97, which is: 0.2611-0.0015=0.2596. b) By using the Central Limit Theorem, the sample mean has a mean of $\mu$ and standard deviation of $\frac{\sigma}{\sqrt n}$. $z_{1}=\frac{value-mean}{standard \ deviation}=\frac{198-205.5}{\frac{8.6}{\sqrt{50}}}=-6.17.$ $z_{1}=\frac{value-mean}{standard \ deviation}=\frac{206-205.5}{\frac{8.6}{\sqrt{60}}}=0.41.$ Using the table, the probability of z being between 0.41 and -6.17 is equal to the probability of z being less than 0.41 minus the probability of z being less than -6.17, which is: 0.6591-0.0001=0.659.
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