## Elementary Statistics (12th Edition)

Mean=$n\cdot p=14 \cdot 0.5=7$. Standard deviation: $\sqrt{n \cdot p \cdot (1-p)}=\sqrt{14 \cdot 0.5 \cdot 0.5}=1.87.$ If a value is unusual, then it is more than two standard deviations far from the mean. $Minimum \ usual \ value=mean-2\cdot(standard \ deviation)=7-2\cdot1.87=3.26.$ $Maximum \ usual \ value=mean+2\cdot(standard \ deviation)=7+2\cdot1.87=10.74$.