Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 8 - Section 8.5 - Normal Distributions - Exercises - Page 603: 16

Answer

$0.866383$

Work Step by Step

Our aim is to find the probability of normal distribution given the mean and standard deviation. Write the equation for standard normal curve. $P(a \leq Z\leq b)=\int_a^b \exp(\dfrac{-t^2}{2}) \ dt$ where, $Z=\dfrac{x-\mu}{\sigma}$ Plug in the above equation the given values to obtain: $P(-1.5 \leq Z \leq 1.5 )=\dfrac{1}{\sqrt 2 \pi}\int_{-1.5}^{1.5} \exp(\dfrac{-t^2}{2}) \ dt$ or, $P(-1.5 \leq Z \leq 1.5 )=0.866383$
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