## Elementary Statistics (12th Edition)

By using the formula: $P(10)=\frac{\mu^k\cdot e^{-\mu}}{x!}=\frac{8.5^{10}\cdot e^{-8.5}}{10!}=0.11.$ If a value is unusual, then it is more than two standard deviations far from the mean. $Minimum \ usual \ value=mean-2\cdot(standard \ deviation)=8.5-2\cdot\sqrt{8.5}=2.67$ $Maximum \ usual \ value=mean+2\cdot(standard \ deviation)=8.5+2\cdot\sqrt{8.5}=14.33$. 10 is between the two bounds therefore it is not unlikely.