Statistics: Informed Decisions Using Data (4th Edition)

Published by Pearson
ISBN 10: 0321757270
ISBN 13: 978-0-32175-727-2

Chapter 6 - Review - Review Exercises - Page 356: 6f

Answer

Expected value: $E(X)=50$ visitors. Standard deviation: $σ_X=6.892$ visitors.

Work Step by Step

n = 1000 and p = 0.05 Expected value: $E(X)=μ_X=np=1000\times0.05=50$ Standard deviation: $σ_X=\sqrt {np(1-p)}=\sqrt {1000\times0.05(1-0.05)}=\sqrt {1000\times0.05\times0.95}=6.892$
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