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: 9

Answer

0.2743.

Work Step by Step

q=1-p=1-0.78=0.22 $n\cdot p=100\cdot 0.22=22\geq5.$ $n\cdot q=100\cdot 0.78=78\geq5.$ Hence, the requirements are satisfied. mean: $\mu=n\cdotp=100\cdot0.22=22$ standard deviation: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{100 \cdot0.22\cdot0.78}=\sqrt{6}=4.14.$ 19.5 is the first one less than 20, hence: $z=\frac{value-mean}{standard \ deviation}=\frac{19.5-22}{4.14}=-0.604.$ By using the table, the probability belonging to z=-0.604: 0.2743.
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