Jose Luis Rodriguez Martinez N11404761 Stochastic Modelling PST 2 Outpatient clinics are hospital clinics that have appointments for patients who visit the hospital without being admitted. These are quite similar to your local General Practitioner’s (GP’s) office. Outpatient clinics at one local hospital include Urology, Spasticity Management, Sex & Fertility, Upper Limb Clinic, Plastic Surgery and Adult spina bifida. At one particular clinic, it is known that the distribution of appointment times (in minutes) looks like 𝑓(𝑡) = 𝑎𝑡 2 + 𝑏𝑡 Where 0 ≤ t ≤ 10 Hospital administrative staff do not know the values of a and b. They do know that the average appointment duration is 5 minutes i. What are the values of a and b? SOLUTION So the first thing we notice that the function is continuous so we can get a valid probability density function (pdf), so we calculate the integral of the function: 10 1 1 10 ∫0 𝑎𝑡 2 + 𝑏𝑡 𝑑𝑡 = 3 𝑎𝑡 3 + 2 𝑏𝑡 2 | 0 = 𝑎103 3 + 𝑏102 2 But remember that the valid pdf is equal to 1 𝑎103 3 + 𝑏102 2 =1 then 𝑎 = 6−300𝑏 2000 Now we got that E[t]= 5 mins Then we must calculate the expected value of a continuous distribution 𝐸[𝑇] = ∫ 𝑡𝑓(𝑡)𝑑𝑡 Thus 10 ∫ 𝑎𝑡 3 + 𝑏𝑡 2 𝑑𝑡 = 0 So we got the average equal to 5, thus 1 4 1 3 10 𝑎104 𝑏103 𝑎𝑡 + 𝑏𝑡 | = + 4 3 4 3 0 𝑎104 𝑏103 60 − 4000𝑏 + = 5 𝑡ℎ𝑒𝑛 𝑎 = 4 3 30000 Now we have to do the substitution: 6 − 300𝑏 60 − 4000𝑏 = 2000 30000 6000 = 1000000𝑏 𝒃= 𝟑 𝟓𝟎𝟎 Now we calculate a value: 𝑎= 6 − 300𝑏 2000 3 6 − 300( ) 500 𝑎= 2000 𝒂= 𝟐𝟏 𝟏𝟎𝟎𝟎𝟎 Question 2 (4 marks) In the Operating Theatre (OT), there are a variety of cancellations that can occur: staff, patient, and equipment based. Unfortunately, short-notice patient based cancellations are the most common. Each patient has a probability of approximately 8% of cancelling on the day of surgery, independently of other patients. i) If the Gastroenterology surgeons have scheduled 8 patients to treat, what is the probability that there will be at most two cancellations? ANSWER This question satisfice the conditions of binomial distribution: • Binary: The possible outcomes of each trial can be classified as a success or a failure. • Independent: Trials must be independent. That is, knowing the outcome of one trial must not tell us anything about the outcome of any other trial. • Number: The number of trials n of the chance process must be fixed in advance. • Same Probability: There is the same probability of success p on each trial So Let X be the number of cancellations on a day n=8 p=0.08 𝑛 𝑛 (𝑥 + 𝑎) = ∑ ( ) 𝑝 𝑥 (1 − 𝑝)𝑛−𝑘 𝑥 𝑛 𝑘=0 so we need to calculate 𝑃(𝑋 ≤ 2) where X ∼ Bin(8, 0.08) 𝑃(𝑋 ≤ 2) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) 8 8 8 𝑃(𝑋 ≤ 2) = ( ) 𝑝0 (1 − 𝑝)8−0 + ( ) 𝑝1 (1 − 𝑝)8−1 + ( ) 𝑝2 (1 − 𝑝)8−2 0 1 2 8 8 8 = ( ) 0.080 (0.92)8−0 + ( ) 0.081 (0.92)8−1 + ( ) 0.082 (0.92)8−2 0 1 2 = 0.5132 + 0.3570 + 0.1086 = 0.9788 ii) For a given surgeon, how many patients do they expect to treat before a patient cancels? For this part, we got that 𝐸(𝑋) = 𝑛𝑝 for a Binomial distribution 𝐸(𝑋) = 𝑛𝑝 = 8 × 0.08 = 0.64 Thus the number of patients expected before a patient cancel is 𝑛 − 𝐸(𝑋) = 8 − 0.64 = 𝟕. 𝟑𝟔 iii) If a short notice surgical block becomes available, prospective patients will accept the time slot with a probability of 20%, independently of other patients. Several patients have cancelled and an administrative assistant is attempting to fill four surgical slots (i.e. obtain four short notice patients) by calling patients on the waiting list. What is the probability that the administrative assistant can fill the blocks in the minimum number of calls? Answer If a short notice surgical block becomes available, prospective patients will accept the time slot with a probability of 20%, independently of other patients. What is the probability that the administrative assistant can fill the blocks in the minimum number of calls? The minimum calls that the assistance have to do is 4 calls. Let X be the minimum of calls that the assistance do So we have to calculate 𝑃(𝑋 = 4). The probability to get the slot in one call is 𝑃(𝑋 = 1) and the second call is 𝑃(𝑋 = 2) therefore we have to calculate the intersections of the probabilities 𝑃(minimum of calls that the assistance) = 𝑃(𝑋 = 1) ∩ 𝑃(𝑋 = 2) ∩ 𝑃(𝑋 = 3) ∩ 𝑃(𝑋 = 4) 𝑃(minimum of calls that the assistance) = 0.2 × 0.2 × 0.2 × 0.2 𝑃(minimum of calls that the assistance) = 0.0016 Question 3 (4 marks) At the emergency department in one of the public hospitals in South East Queensland, patients arrive according to a Poisson distribution with an average of 15 minutes between arrivals. i. What is the probability that the hospital experiences a busy period and more than three patients arrive in half an hour? Here we got that in average 15 mins between arrivals, so 𝐸[𝑋] = 𝜆 But I want to calculate 𝜆 In terms of hours, so 𝜆 = 4𝑡 Let X be the numbers of arrivals of patients in t hours, then 𝑋 ~ 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 (𝜆 = 4𝑡) 1 2 Now we want to know the probability in 30 mins, so 𝑡 = 1/2 𝑋 ~ 𝑃𝑜 (𝜆 = 4 ( )) = (𝜆 = 2) Probability that more than 3 patients arrives in a half hour: 𝑃(𝑋 > 3) = 1 − 𝑃(𝑋 ≤ 3) = 1 − [ 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3)] 3 =1−∑ 𝑥=0 1 2 𝑒 −2 2𝑥 𝑥! 4 8 = 1 − (𝑒 −2 (0! + 1! + 2! + 3!))≈ 0.14288 ii) Last Saturday, between the hours of 10am and 12pm, there were 10 emergency arrivals. A nurse on shift-work was required to triage all emergency arrivals between 9am and 1pm that day. How many patients were expected to arrive over that time? Answer Again here we got that 𝐸[𝑋] = 𝜆 . we got 10 emergency arrivals in 2 hours, means 5 in 1 hour; therefore 𝜆 = 5𝑡 Let X be the numbers of emergency arrivals of patients in t hours then 𝑋 ~ 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 (𝜆 = 5𝑡) now the new interval of time is between 9 am to 1pm, so we got 4 hours; means that 𝑡 = 4 thus 𝐸[𝑋] = 𝜆 𝐸[𝑋] = 5𝑡 𝐸[𝑋] = 5(4) = 20 So we expected 20 arrivals between 9am and 1pm. iii) An on-call doctor is attempting to nap in between emergency arrivals. The doctor is going to set their alarm such that the probability of a patient arriving in that time is less than 40%. How long can the doctor set their alarm for? Answer According with last question the emergency arrivals is between 10am to 12pm and 𝐸[𝑋] = 𝜆 with 𝜆 = 5𝑡 Let T be the time arrival (in hours) for which the doctor sets the alarm. then 𝑋 ~ 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 (𝜆 = 5𝑡) so the probability of a patient arrives at that time 𝑃(𝑋 ≥ 1) 𝑃(𝑋 ≥ 1) = 1 − 𝑃(𝑋 = 0) 𝜆 𝑃(𝑋 ≥ 1) = 1 − ∑ 𝑥=0 𝜆 𝑃(𝑋 ≥ 1) = 1 − ∑ 𝑥=0 =1− 𝑒 −𝜆 𝜆𝑥 𝑥! 𝑒 −5𝑡 5𝑡 𝑥 𝑥! 𝑒 −5𝑡 5𝑡 0 < 0.40 0! 𝑒 −5𝑡 5𝑡 0 ≥ 0.60 0! 𝑒 −5𝑡 ≥ 0.60 −5𝑡 ln 𝑒 ≥ ln 0.60 −5𝑡 ≥ ln 0.60 𝑡≤− ln 0.60 5 𝑡≤− (−0.5108) 5 𝑡≤ 0.5108 5 𝑡 ≤ 0.10216 ℎ𝑜𝑢𝑟𝑠 𝑡 ≤ 6.1296 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 The maximum time for the alarm to be set is 6.1296 𝑚𝑖𝑛𝑢𝑡𝑒𝑠