Answer
Mean:12. Standard deviation:3.1. Minimum usual value:5.8, maximum usual value:18.2.
Work Step by Step
Mean=$n\cdot p=60 \cdot 0.2=12$.
Standard deviation: $\sqrt{n \cdot p \cdot (1-p)}=\sqrt{60 \cdot 0.2 \cdot 0.8}=3.1.$
If a value is unusual, then it is more than two standard deviations far from the mean. $Minimum \ usual \ value=mean-2\cdot(standard \ deviation)=12-2\cdot3.1=5.8$
$Maximum \ usual \ value=mean+2\cdot(standard \ deviation)=12+2\cdot3.1=18.2$.
