Answer
The distribution is approximately normal with mean of 0.72 and $σ_{p̂} = 0.018$
Work Step by Step
np(1-p) = 600 x 0.72 x 0.28 = 120.96 $\geq 10$
$μ_{p̂} = p = 0.72$
$σ_{p̂} = \sqrt \frac{p(1-p)}{n}
= \sqrt \frac{.72(.28)}{600} = 0.018$
Statistics: Informed Decisions Using Data (4th Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0321757270
ISBN 13:
978-0-32175-727-2
Chapter 8 - Review - Review Exercises - Page 422: 8a
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