Elementary Statistics (12th Edition)

Published by Pearson
ISBN 10: 0321836960
ISBN 13: 978-0-32183-696-0

Chapter 6 - Normal Probability Distributions - 6-7 Normal as Approximation to Binomial - Basic Skills and Concepts: 19

Answer

The observed rate seems to be very unlikely.

Work Step by Step

p=0.61 $q=1-p=1-0.61=0.39$ $n⋅p=1002⋅0.61=611.22≥5.$ $n⋅q=1002⋅0.39=390.78≥5.$ Hence, the requirements are satisfied. mean: $\mu=n\cdotp=1002\cdot0.61=611.22.$ standard deviation: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{1002\cdot0.61\cdot0.39}=15.44.$ 700.5 is the first value lower than 701, hence: $z=\frac{value-mean}{standard \ deviation}=\frac{700.5-611.22}{15.44}=5.78.$ By using the table, the probability belonging to z=5.78: 0.9999, hence the probability: 1-0.9999=0.0001. This probability is really close to 0, therefore the observed rate seems to be very unlikely.
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