## Introductory Statistics 9th Edition

Published by Wiley

# Chapter 6 - Section 6.1 - Continuous Probability Distribution and the Normal Probability Distribution - Exercises - Page 241: 6.16

#### Answer

a.$P(-1.83 \leq z \leq 2.57 )$ = .9949-.0336=.9613 b. $P(0 \leq z \leq 2.02 )$ =.9783-.5=.4783 c.$P(-1.99 \leq z \leq 0 )$ = .5-.0233 = .4767 d. $P(z\geq 1.48 )$ = 1-$P(z\leq 1.48 )$ = 1- 0.9306 = .0694

#### Work Step by Step

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