Essentials of Statistics (5th Edition)

Published by Pearson
ISBN 10: 0-32192-459-2
ISBN 13: 978-0-32192-459-9

Chapter 6 - Normal Probability Distributions - 6-7 Normal as Approximation to Binomial - Page 306: 10



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: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{100\cdot0.22\cdot0.78}=\sqrt{3.12}=4.1425.$ 24.5 is the first one less than 25, hence: $z=\frac{value-mean}{standard \ deviation}=\frac{24.5-22}{4.1425}=0.6.$ By using the table, the probability belonging to z=0.6: 0.7257, hence the probability of z being more than 0.6: 1-0.7257=0.2743.
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