# Answer in Statistics and Probability for Aisha #204899

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).

Let “X=” the average number of patients arriving per day: “Xsim Po(lambda).”

“P(X=x)=dfrac{e^{-lambda}cdotlambda^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”