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 311: 8



Work Step by Step

q=1-p=1-0.4=0.6 $n\cdot p=25\cdot 0.4=10\geq5.$ $n\cdot q=25\cdot 0.6=15\geq5.$ Hence, the requirements are satisfied. mean: $\mu=n\cdotp=25\cdot0.4=10$ standard deviation: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{25 \cdot0.4\cdot0.6}=\sqrt{6}=2.45.$ 9.5 is the first one more than 9, hence: $z=\frac{value-mean}{standard \ deviation}=\frac{9.5-10}{2.45}=-0.2.$ By using the table, the probability belonging to z=-0.2: 0.4207, hence the probability of z being more than 0.2: 1-0.4207=0.5793.
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