Elementary Statistics (12th Edition)

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

Chapter 6 - Normal Probability Distributions - 6-3 Applications of Normal Distributions - Basic Skills and Concepts - Page 268: 16



Work Step by Step

$z_1=\frac{value-mean}{standard \ deviation}=\frac{120-100}{15}=1.333.$ $z_2=\frac{value-mean}{standard \ deviation}=\frac{110-100}{15}=0.667.$ Using the table, the value belonging to 1.333: 0.9082, the value belonging to 0.667: 0.7486. 0.9082-0.7486=0.1596.
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.