The number of cases of tetanus reported in the US in a single month has...

Let X be the number of cases in 1 month
X follows Poisson distribution with parameter (rate) ${\displaystyle \lambda =4}$
${\displaystyle X\sim }$ Poisson ${\displaystyle (\lambda =4)}$
${\displaystyle P(X\ge 5),\lambda =4}$
${\displaystyle P(X\ge 5)=1-P(0\le X\le 4)}$ where ${\displaystyle P(0\le X\le 4)=P(X=0)+P(X=1)}$
+P(X=2)+P(X=3)+P(X=4)
${\displaystyle P(X=0)=\frac{e^{-\lambda }{\lambda }^0}{0!}=\frac{e^{-4}{\left(4\right)}^0}{0!}=0.018316}$
${\displaystyle P(X=1)=\frac{e^{-\lambda }{\lambda }^1}{1!}=\frac{e^{-4}{\left(4\right)}^1}{1!}=0.073263}$
${\displaystyle P(X=2)=\frac{e^{-\lambda }{\lambda }^2}{2!}=\frac{e^{-4}{\left(4\right)}^2}{2!}=0.146525}$
${\displaystyle P(X=3)=\frac{e^{-\lambda }{\lambda }^3}{3!}=\frac{e^{-4}{\left(4\right)}^3}{3!}=0.195367}$
${\displaystyle P(X=4)=\frac{e^{-\lambda }{\lambda }^4}{4!}=\frac{e^{-4}{\left(4\right)}^4}{4!}=0.195367}$
Answer=1- (0.018316 + 0.073263 + 0.146525 + 0.195367 + 0.195367)=0.371163
Use excel function POISSON.DIST (X, mean, cumulative) to calculate
${\displaystyle P(X\ge 5)=1-P(X\le 4)=1-F_X\left(4\right):}$
1-POISSON.DIST (4,4 TRUE)=0.37116306482
${\displaystyle P(X\ge 5)=0.3712}$