Introductory Statistics 9th Edition

Published by Wiley
ISBN 10: 1-11905-571-7
ISBN 13: 978-1-11905-571-6

Chapter 6 - Section 6.2 - Standardizing a Normal Distribution - Exercises - Page 247: 6.23

Answer

a. From the given normal distribution, μ= 117.6 and σ= 14.6 . For x =77.9 : z = $\frac{77.9-117.6}{14.6}$ = -2.7192 For x =98.3 : z = $\frac{98.3-117.6}{14.6}$ = -1.322 As shown in the figure, the required area is given by the area under the standard normal distribution curve between z=-2.7192 and -1.322 . This area is approximately 0.898. P($77.9 \leq x \leq 98.3$) = P($ -2.7192\leq z \leq -1.322$ ) = .898 approximately. b. For x =85.3 : z = $\frac{85.3-117.6}{14.6}$ = -2.2123 For x =142.6 : z = $\frac{142.6-117.6}{14.6}$ = 1.7123 As shown in the figure, the required area is given by the area under the standard normal distribution curve between z= -2.2123 and 1.7123 . This area is approximately 0.9431. $P(85.3 \leq x \leq 142.6)$ = $P(-2.2123 \leq z \leq 1.7123 )$ = .9431 approximately.
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Work Step by Step

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