## Elementary Statistics (12th Edition)

Published by Pearson

# Chapter 6 - Normal Probability Distributions - 6-7 Normal as Approximation to Binomial - Basic Skills and Concepts - Page 312: 10

0.2005.

#### 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: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{100\cdot0.22\cdot0.78}=\sqrt{3.12}=4.14.$ 25.5 is the first one more than 25, hence: $z=\frac{value-mean}{standard \ deviation}=\frac{25.5-22}{4.14}=0.84.$ By using the table, the probability belonging to z=0.84: 0.7995, hence the probability of z being more than 0.84: 1-0.7995=0.2005.

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