Answer
a) Mean:20. Standard deviation:4.
b)It is not unusually high. Also, the rate seems to be correct.
Work Step by Step
a) Mean=$n\cdot p=100 \cdot 0.2=20$.
Standard deviation: $\sqrt{n \cdot p \cdot (1-p)}=\sqrt{100 \cdot 0.2 \cdot 0.8}=4.$
b) If a value is unusual, then it is more than two standard deviations far from the mean. $Minimum \ usual \ value=mean-2\cdot(standard \ deviation)=20-2\cdot4=12$
$Maximum \ usual \ value=mean+2\cdot(standard \ deviation)=20+2\cdot4=28$.
25 is between these two values, therefore it is not unusually high. Also, the rate seems to be correct.
