Answer
a) Mean:0.0821918. Standard deviation:0.2862981.
b)It is unusually high.
Work Step by Step
Here, n=30 and p=$\frac{1}{365}$.
a) Mean=$n\cdot p=30 \cdot \frac{1}{365}=0.0821918$.
Standard deviation: $\sqrt{n \cdot p \cdot (1-p)}=\sqrt{30 \cdot \frac{1}{365} \cdot \frac{364}{365}}=0.2862981.$
b) If a value is unusual, then it is more than two standard deviations far from the mean. $Minimum \ usual \ value=mean-2\cdot(standard \ deviation)=0.0821918-2\cdot0.2862981=-0.4904044$
$Maximum \ usual \ value=mean+2\cdot(standard \ deviation)=0.0821918+2\cdot0.2862981=0.654788$.
2 is more than the upper bound, therefore it is unusually high.