Deragz

2022-01-18

The number of cases of tetanus reported in the US in a single month has a Poisson distribution with a parameter of $\lambda =4.0$. What is the probability that five or more cases will be reported?

Lindsey Gamble

Expert

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

Do you have a similar question?