Question

At a particular trauma center in Queens, NY, on average there are 2.0 blunt force trauma admissions and 0.5 penetrating trauma admissions per day (per 24 hours).

a) (4 pts) Calculate the probability that there will be exactly 3 blunt force trauma admissions on a randomly selected day.

b) (6 pts) Calculate the probability of 4 or more blunt force trauma admissions on a randomly selected day.

c) (3 pts) How many blunt force trauma admissions would be expected on a randomly selected day?

d) (6 pts) Assume that blunt force trauma admissions and penetrating trauma admissions are independent. What is the probability that there are no blunt force trauma admissions and no penetrating trauma admissions on a randomly selected day? (Hint: Think back to probability rules….)

Answer 1

a)from poisson distribution

probability that there will be exactly 3 blunt force trauma admissions on a randomly selected day

=P(X=3)=e^-22³/3! =0.1804

b) probability of 4 or more blunt force trauma admissions on a randomly selected day

=P(X>=4)=1-P(X<=3)=1-(P(X=0)+P(X=1)+P(X=2)+P(X=3))

=1-(e^-22⁰/0!+e^-22¹/1! +e^-22²/2! +e^-22³/3! )

=0.142877

c)

expected blunt force trauma admissions expected on a randomly selected day =2

d)

probability that there are no blunt force trauma admissions and no penetrating trauma admissions on a randomly selected day =(e^-22⁰/0!)*(e^-0.5*0.5⁰/0!)=0.0821