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: 21


0.1075, evidence is not too strong.

Work Step by Step

p=0.2 $q=1-p=1-0.2=0.8$ $n⋅p=50⋅0.2=10≥5.$ $n⋅q=50⋅0.8=40≥5.$ Hence, the requirements are satisfied. mean: $\mu=n\cdotp=50\cdot0.2=10.$ standard deviation: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{50\cdot0.2\cdot0.8}=2.83.$ 6.5 is the first value more than 6, hence: $z=\frac{value-mean}{standard \ deviation}=\frac{6.5-10}{2.83}=-1.24.$ By using the table, the probability belonging to z=-1.24: 0.1075, hence the probability: 0.1075. This probability is not too close to 0, therefore the evidence is not very strong.
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