## Elementary Statistics (12th Edition)

Published by Pearson

# Chapter 6 - Normal Probability Distributions - 6-7 Normal as Approximation to Binomial - Basic Skills and Concepts: 18

#### Answer

Observed rate seems to be very unlikely.

#### Work Step by Step

p=0.85 $q=1-p=1-0.85=0.15$ $n⋅p=523⋅0.85=444.55≥5.$ $n⋅q=523⋅0.15=78.45≥5.$ Hence, the requirements are satisfied. mean: $\mu=n\cdotp=523\cdot0.85=444.45.$ standard deviation: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{523\cdot0.85\cdot0.15}=8.17.$ 517.5 is the first value lower than 518, hence: $z=\frac{value-mean}{standard \ deviation}=\frac{517.5-444.45}{8.17}=8.93.$ By using the table, the probability belonging to z=8.93: 0.9999, hence the probability: 1-0.9999=0.0001. This probability is really close to 0, therefore the observed rate seems to be very unlikely.

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