Poisson Analysis: Asthma ED Visits in Seattle

Step 1 of 4:

Given for an interval of the length of time T with the parameter \(\lambda\).

From the given information the mean number of asthma emergency department visit per day is 1.8.

\(\lambda=1.8\).

Let \(\mathrm{X}\) follows the number of visit in a day.

Then the poisson variable \(\mathrm{X}\)’s probability function is

\(\mathrm{P}(\mathrm{X}=\mathrm{x})=e^{-\lambda T} \frac{(\lambda T)^{x}}{x !}\)

a). Now we have to find the probability of more than 5 visits in a day.

Then the probability of more than 5 visits in a day is

\(\mathrm{P}(\mathrm{X}>5)=1-[\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)+, \ldots,+\mathrm{P}(\mathrm{X}=5)]\)

We know that \(\lambda T=1.8\).

\(\mathrm{P}(\mathrm{X}>5)=1-\left(e^{-1.8} \frac{(1.8)^{0}}{0 !}+e^{-1.8} \frac{(1.8)^{1}}{1 !}+e^{-1.8} \frac{(1.8)^{2}}{2 !}+e^{-1.8} \frac{(1.8)^{3}}{3 !}+e^{-1.8} \frac{(1.8)^{4}}{4 !}+e^{-1.8} \frac{(1.8)^{5}}{5 !}\right)\)

\(\mathrm{P}(\mathrm{X}>5)=1-\left\{e^{-1.8} \frac{(1.8)^{0}}{0 !}+e^{-1.8} \frac{(1.8)^{1}}{1}+e^{-1.8} \frac{(1.8)^{2}}{2 \times 1}+e^{-1.8} \frac{(1.8)^{3}}{3 \times 2 \times 1}+e^{-1.8} \frac{(1.8)^{4}}{4 \times 3 \times 2 \times 1}+e^{-1.8} \frac{(1.8)^{5}}{5 \times 4 \times 3 \times 2 \times 1}\right\}\)

\(P(X>5)=1-(0.165298+0.297537+0.267784+0.1606705+0.072301+0.02602)\)        

\(\mathrm{P}(\mathrm{X}>5)=1-0.9896\)

\(P(X>5)=0.0104\)

Therefore, the probability of more than 5 visits in a day is 0.0104.