# Chapter 5 - Discrete Probability Distributions - Review - Review Exercises - Page 237: 3

Mean:240, standard deviation:12, it is not unusually low or high.

#### Work Step by Step

Here n=600, p=0.4. Mean=$n\cdot p=600 \cdot 0.4=240$. Standard deviation: $\sqrt{n \cdot p \cdot (1-p)}=\sqrt{600 \cdot 0.4 \cdot 0.6}=12.$ If a value is unusual, then it is more than two standard deviations far from the mean. $Minimum \ usual \ value=mean-2\cdot(standard \ deviation)=240-2\cdot12=216$ $Maximum \ usual \ value=mean+2\cdot(standard \ deviation)=240+2\cdot12=264$. Our value is between these two values, therefore it is not unusually low or high.

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