Question

The number of people arriving for treatment at an emergency room can be modeled by aPoisson process with a rate parameter of 5/hr.(a) What is the probability thatexactly fourarrivals occur during a particular hour?(b) What is the probability thatat least threepeople arrive during a particular hour?(c) How many people do you expect to arrive during a45-minuteperiod?

Answer 1

Answer:

(a) The probability of having exactly four arrivals during a particular hour is 0.1754.

(b) The probability that at least 3 people arriving during a particular hour is 0.7350.

(c) The expected arrivals in a 45 minute period (0.75 hours) is 3.75 arrivals.

Step-by-step explanation:

(a) If the arrivals can be modeled by a Poisson process, with λ = 5/hr, the probability of having exactly four arrivals during a particular hour is:

$P(X=4)=\frac{\lambda^{X}*e^{-\lambda} }{X!} =\frac{5^{4}*e^{-5} }{4!}=\frac{625*0.006737947}{24} =\frac{4.211}{24}=0.1754$

The probability of having exactly four arrivals during a particular hour is 0.1754.

(b) The probability that at least 3 people arriving during a particular hour can be written as

$P(X>3)=1-P(X\leq3)=1- (P(0)+P(1)+P(2)+P(3))$

Using

$P(X)=\frac{\lambda^{X}*e^{-\lambda} }{X!}$

We get

$P(X>3)=1-P(X\leq3)=1- (P(0)+P(1)+P(2)+P(3))\\P(X>3)=1-(0.0067+ 0.0337+ 0.0842 + 0.1404 )\\P(X>3)=1-0.2650=0.7350$

The probability that at least 3 people arriving during a particular hour is 0.7350.

(c) The expected arrivals in a 45 minute period (0.75 hours) is

$EV=\lambda*t=5*0.75=3.75$