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: 11

Answer

0.0928.

Work Step by Step

$q=1-p=1-0.78=0.22$ $n⋅p=100⋅0.22=22≥5.$ $n⋅q=100⋅0.78=78≥5.$ Hence, the requirements are satisfied. mean: $\mu=n\cdotp=100\cdot0.22=22.$ standard deviation: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{100\cdot0.22\cdot0.78}=\sqrt{3.12}=4.14.$ 23 is between 23.5 and 22.5, hence: $z_{1}=\frac{value-mean}{standard \ deviation}=\frac{22.5-22}{4.14}=0.12.$ $z_{2}=\frac{value-mean}{standard \ deviation}=\frac{23.5-22}{4.14}=0.36.$ By using the table, the probability belonging to z=0.36: 0.6406, to z=0.12: 0.5478, hence the probability: 0.6406-0.5478=0.0928.
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