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 - Page 313: 24


0.1736, the evidence is not strong.

Work Step by Step

p=0.67 $q=1-p=1-0.67=0.33$ $n⋅p=250⋅0.67=167.5≥5.$ $n⋅q=250⋅0.33=82.5≥5.$ Hence, the requirements are satisfied. mean: $\mu=n\cdotp=250\cdot0.67=167.5.$ standard deviation: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{250\cdot0.67\cdot0.33}=7.43.$ 174.5 is the first value less than 70%, hence: $z=\frac{value-mean}{standard \ deviation}=\frac{174.5-167.5}{7.43}=0.94.$ By using the table, the probability belonging to z=0.94: 0.8264, hence the probability: 1-0.8264=0.1736. This probability is not really close to 0, hence the evidence is not strong.
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