Respuesta :
Answer:
a) P(x=3)=0.089
b) P(x≥3)=0.938
c) 1.5 arrivals
Step-by-step explanation:
Let t be the time (in hours), then random variable X is the number of people arriving for treatment at an emergency room.
The variable X is modeled by a Poisson process with a rate parameter of λ=6.
The probability of exactly k arrivals in a particular hour can be written as:
[tex]P(x=k)=\lambda^{k} \cdot e^{-\lambda}/k!\\\\P(x=k)=6^k\cdot e^{-6}/k![/tex]
a) The probability that exactly 3 arrivals occur during a particular hour is:
[tex]P(x=3)=6^{3} \cdot e^{-6}/3!=216*0.0025/6=0.089\\\\[/tex]
b) The probability that at least 3 people arrive during a particular hour is:
[tex]P(x\geq3)=1-[P(x=0)+P(x=1)+P(x=2)]\\\\\\P(0)=6^{0} \cdot e^{-6}/0!=1*0.0025/1=0.002\\\\P(1)=6^{1} \cdot e^{-6}/1!=6*0.0025/1=0.015\\\\P(2)=6^{2} \cdot e^{-6}/2!=36*0.0025/2=0.045\\\\\\P(x\geq3)=1-[0.002+0.015+0.045]=1-0.062=0.938[/tex]
c) In this case, t=0.25, so we recalculate the parameter as:
[tex]\lambda =r\cdot t=6\;h^{-1}\cdot 0.25 h=1.5[/tex]
The expected value for a Poisson distribution is equal to its parameter λ, so in this case we expect 1.5 arrivals in a period of 15 minutes.
[tex]E(x)=\lambda=1.5[/tex]
