3. A turnstile counts people entering a small museum, and obtains the following

Question

3. A turnstile counts people entering a small museum, and obtains the following data for the last few Wednesdays. The museum is open from9:00am to 9:00pm Wednesday 13 /10 1/17 1/24 131 272/14 2/212/28 Number of people 218 196 195 200 195 204 223 201 208 Use a Poisson process to model people arriving to the museum on a Wednesday during open hours a) What should 2 be? (Hint: for an average rate, use the average of the data, and check the units.) b) What is the expected value and standard deviation of the amount of time, in minutes, that will pass between any two persons' arrivals? c) What is the probability that at least 5 minutes will pass between any two persons' arrivals? d) What is the probability that exactly one person will go through the turnstile during any fifteen- minute interval? e) What is the probability that at least one person will go through the tunstile during any fifteen- minute interval? f Right now, the count is at 100. What is the chance that the count will be at 120 in one hour?

Explanation / Answer

(a) Average of the 9 counts = 204.44

So on an average, 204.44 people enter the museum per day in a 12 hour interval.

We can model the Poisson distribution on an hourly basis

= number of people arriving per hour = 204.44.12 = 17.04

(b) The interarrival times are exponentially distributed.

Expected value of the exponential distribution = 1/ = 0.058 hour = 3.52 minutes

Standard deviation of the interarrival time = 1/2 = 12.4 minute2

(c) For the exponential distribution, we need to calculate P(X>5/60),i.e., P(X>0.0833)

P(X>0.0833) = 1-P(X0.0833) = 1- F(0.0833) = 1-[1-e-*0.0833] = e-*0.0833 = 0.2418

(d) For the Poisson distribution, we need to find P(k=1), with *t = 17.04 * (15/60) = 4.259

P(k=1) = e-4.259 * (4.259)1 / 1! = 0.0602

(e) P(k=0) = e-4.259 * (4.259)0 / 0! = 0.0141

P(k1) = 1-0.0141 = 0.9859

(f) P(k=20) = e-17.04 * (17.04)20 / 20! = 0.0696

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