Answer to Question #204899 in Statistics and Probability for Aisha

Question

In a hospital’s emergency for Covid 19 patients, the average number of patients arriving 

between 9:00 a.m to 6:00 p.m is 7 per day. Find the probability that, on a given day, the 

number of patients arriving at the emergency room will be exactly (R + 1).

Answer 1

Let $X=$ the average number of patients arriving per day: $X\sim Po(\lambda).$

$$P(X=x)=\dfrac{e^{-\lambda}\cdot\lambda^x}{x!}$$

Given $\lambda=7$

$$P(X=R+1)=\dfrac{e^{-7}\cdot 7^{R+1}}{(R+1)!}$$

$$R=0:P(X=0+1)=\dfrac{e^{-7}\cdot 7^{0+1}}{(0+1)!}\approx0.00638$$

$$R=1:P(X=1+1)=\dfrac{e^{-7}\cdot 7^{1+1}}{(1+1)!}\approx0.02234$$

$$R=2:P(X=2+1)=\dfrac{e^{-7}\cdot 7^{2+1}}{(2+1)!}\approx0.05213$$

$$R=3:P(X=3+1)=\dfrac{e^{-7}\cdot 7^{3+1}}{(3+1)!}\approx0.09123$$

$$R=4:P(X=4+1)=\dfrac{e^{-7}\cdot 7^{4+1}}{(4+1)!}\approx0.12772$$

$$R=5:P(X=5+1)=\dfrac{e^{-7}\cdot 7^{5+1}}{(5+1)!}\approx0.14900$$

$$R=6:P(X=6+1)=\dfrac{e^{-7}\cdot 7^{6+1}}{(6+1)!}\approx0.14900$$

$$R=7:P(X=7+1)=\dfrac{e^{-7}\cdot 7^{7+1}}{(7+1)!}\approx0.13038$$

$$R=8:P(X=0+1)=\dfrac{e^{-7}\cdot 7^{8+1}}{(8+1)!}\approx0.10140$$

$$R=9:P(X=9+1)=\dfrac{e^{-7}\cdot 7^{9+1}}{(9+1)!}\approx0.07098$$