Answer
$412.68,16.4548$
Work Step by Step
The mean of a binomial distribution: $\mu=np=1200\cdot0.3439=412.68$
The standard deviation of a binomial distribution: $\sigma=\sqrt{npq}=\sqrt{np(1-p)}=\sqrt{1200\cdot0.3439\cdot(1-0.3439)}=16.4548$
The Basic Practice of Statistics 7th Edition
by Moore, David S.; Notz, William I.; Fligner, Michael A.
Published by
W. H. Freeman
ISBN 10:
146414253X
ISBN 13:
978-1-46414-253-6
Chapter 14 - Binomial Distributions - Chapter 14 Exercises - Page 342: 14.33b
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