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rooman

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Jun 6, 2022
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Consider a large walk-in (i.e. no appointment) General Practice clinic for which
the number of people arriving for a consultation between 2 and 3pm follows a Pois-
son distribution with parameter λ = 20. Patients present to the clinic with one
complaint to speak to the doctor about, and others present with two com-
pliants. [ assume nobody presents with more than 2 complaints!] Assume
patients persent with one complaint with probability 0.70, and with 2 com-
plaints with probability 0.30, and all patients are independent.
What is the expected total number of issues/complaints across all patients presenting
to the clinic between 2-3pm on a given day?
 
Consider a large walk-in (i.e. no appointment) General Practice clinic for which
the number of people arriving for a consultation between 2 and 3pm follows a Pois-
son distribution with parameter λ = 20. Patients present to the clinic with one
complaint to speak to the doctor about, and others present with two com-
pliants. [ assume nobody presents with more than 2 complaints!] Assume
patients persent with one complaint with probability 0.70, and with 2 com-
plaints with probability 0.30, and all patients are independent.
What is the expected total number of issues/complaints across all patients presenting
to the clinic between 2-3pm on a given day?
Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem
 
consider X be the number of people arriving between 2 and 3pm and Y be the total
number of complaints so E(Y | X = x), stuck getting value of x
 
consider X be the number of people arriving between 2 and 3pm and Y be the total
number of complaints so E(Y | X = x), stuck getting value of x
\(\displaystyle E[\text{complaints]}=E[\text{number of patients}]\times E[\text{complaints per patient}]\)
\(\displaystyle \text{Number of patients}\sim\text{Poisson}(\lambda=20)\)
\(\displaystyle \text{Complaint(s) per patient}=\begin{cases} 1 &\Pr(\text{one complaint}) = 0.7 \\ 2 &\Pr(\text{two complaints})=0.3 \end{cases}\)

Find their respective expected values and multiply.
 
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