Answer
a) Mean:0.000013. Standard deviation:0.003649.
b)It is unusual.
Work Step by Step
Here, n=$2600, p=\frac{1}{195249054}.$
a) Mean=$n\cdot p=2600 \cdot \frac{1}{195249054}=0.000013$.
Standard deviation: $\sqrt{n \cdot p \cdot (1-p)}=\sqrt{2600 \cdot \frac{1}{195249054} \cdot \frac{195249053}{195249054}}=0.003649.$
b) If a value is unusual, then it is more than two standard deviations far from the mean. $Minimum \ usual \ value=mean-2\cdot(standard \ deviation)=0.000013-2\cdot0.003649=-0.007285$
$Maximum \ usual \ value=mean+2\cdot(standard \ deviation)=0.000013+2\cdot0.003649=0.007311$.
1 is more than the upper bound, therefore it is unusual.