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: 12

Answer

0.0794

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: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{100\cdot0.22\cdot0.78}=\sqrt{3.12}=4.14.$ 19 is between 19.5 and 18.5, hence: $z_{1}=\frac{value-mean}{standard \ deviation}=\frac{19.5-22}{4.14}=-0.6.$ $z_{2}=\frac{value-mean}{standard \ deviation}=\frac{18.5-22}{4.14}=-0.84.$ By using the table, the probability belonging to z=-0.6: 0.2743, to z=-0.84: 0.1949, hence the probability: 0.2743-0.1949=0.0794.
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