An emergency room at a particular hospital gets an average of five patients per hour. A doctor wants to know the probability that the ER gets more than five patients per hour. Give the reason why this would be a Poisson distribution.

An emergency room at a particular hospital gets an average of five patients per hour. A doctor wants to know the probability that the ER gets more than five patients per hour.

Let X be the number patients per hour the ER gets in the hospital. Therefore the sample size X takes on the values $0,1,2,3,4,5,6,7,8,9,10\dots .$. This is a poisson distribution as the events occurring in a fixed interval of time one hours and independently occur since the last event.

Here X follows a poisson distribution with the probability mass function of poisson distribution as

$P(X=x)=\frac{e^{-\mu }{\mu }^x}{x!}$

According to the question, the average number of five patients per hour in an emergency room. So average of Poisson distribution is $\mu =5$.

Therefore the probability that the ER gets more than five patients per hour is determined as:

$\begin{array}{r}P(X>5)=1-P(X\le 5)\\ =0.3840\end{array}$

Step in TI-83 + calculator as follows:

- Press 1 - and then press $2^{\text{nd}}$DISTR

- Arrow down to poissoncdf. Press ENTER

- Enter$(5,5)$

- The result shows

Therefore the required probability that the ER gets more than five patients per hour$P(X>5)=0.3840.$