# Chapter 5 - Discrete Probability Distributions - 5-4 Parameters for Binomial Distributions - Basic Skills and Concepts - Page 226: 9

a) Mean:145.5. Standard deviation:8.53. b)It is an unusually high number and the method seems to be effective.

#### Work Step by Step

a) Mean=$n\cdot p=291 \cdot 0.5=145.5$. Standard deviation: $\sqrt{n \cdot p \cdot (1-p)}=\sqrt{291 \cdot 0.5 \cdot 0.5}=8.53.$ b) If a value is unusual, then it is more than two standard deviations far from the mean. $Minimum \ usual \ value=mean-2\cdot(standard \ deviation)=145.5-2\cdot8.53=128.56$ $Maximum \ usual \ value=mean+2\cdot(standard \ deviation)=145.5+2\cdot8.53=162.56$. 239 is not between these numbers, therefore it is an unusually high number and the method seems to be effective.

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