Elementary Statistics: Picturing the World (6th Edition)

Published by Pearson
ISBN 10: 0-32191-121-0
ISBN 13: 978-0-32191-121-6

Chapter 5 - Normal Probability Distributions - Section 5.4 Sampling Distributions and the Central Limit Theorem - Exercises - Page 270: 16

Answer

$P(\bar{x} > 24.3) = 0.0082$ It is unusual for the mean to be greater than 24.3

Work Step by Step

n = 100 $\sigma$ = 1.25 $\mu$ =24 Want to find P($\bar{x}$ > 24.3): i) Find the z score corresponding to 24.3 z = $\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt n}}$ z = $\frac{24.3 - 24}{\frac{1.25}{\sqrt 100}}$ z = 2.4 ii) $P( z > 2.4) = 1 - P( z < 2.4)$ = 1 - 0.9918 = 0.0082 iii) Therefore $P(\bar{x} > 24.3) = 0.0082$ It is unusual for the mean to be greater than 24.3. We can infer this because the corresponding z-score is greater than 2.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.