Elementary Statistics (12th Edition)

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.