Answer
Mean:6.016. Standard deviation:2.373. Minimum usual value:1.27, maximum usual value:10.762.
Work Step by Step
Mean=$n\cdot p=94 \cdot 0.064=6.016$.
Standard deviation: $\sqrt{n \cdot p \cdot (1-p)}=\sqrt{94 \cdot 0.064 \cdot 0.936}=2.373.$
If a value is unusual, then it is more than two standard deviations far from the mean. $Minimum \ usual \ value=mean-2\cdot(standard \ deviation)=6.016-2\cdot2.373=1.27$
$Maximum \ usual \ value=mean+2\cdot(standard \ deviation)=6.016+2\cdot2.373=10.762$.
