Tutorial 4

1. Tutorial 4 February 4, 2013

2. Problem 1 At Tony And Cleo’s bakery, one kind of birthday cake is offered. It takes 15 minutes to decorate this particular cake, and the job is performed by one particular baker. In fact, this is all this baker does. What mean time between arrivals (exponentially distributed) can be accepted if the mean length of the queue for decorating is not to exceed 5 cakes?

3. Solution 1 • Model this system as M/G/1 queue (exponential arrivals/generalized service time/single server) • The service rate is =4/hour, and variance 2=0 • For this queue, we know LQ = Long-run average time spent in system per customer is given by: LQ = 2 (1+ 2  2)/2(1- ), where  2 =0 and = λ/ • Hence, LQ =λ2/{2  2(1-[λ/])}

4. Solution 1 • We need to find λ such that LQ does not exceed 5 • We rearrange the equation, and find: Since only positive values of λ makes sense, Using LQ = 5 and  =4, we get λ = 3.6634

5. Problem 2 Records pertaining to the monthly number of job-related injuries at an underground coal mine were being studied by a federal agency. The values for the past 100 months were as follows: Apply the chi-square test to these data to test the hypothesis that the underlying distribution is Poisson. Use a level of significance of =.05.

6. Solution 2 • Find mean • Find probability density for the Poisson distribution in which mean is known • 2 should not be calculated if the expected value in any category is less than 5. Hence we group a subset x =[3..6] and re-evaluate the table to estimate(oi-Ei)2/Ei

7. Solution 2 • (Oi-Ei)2/Ei = 3.411; 2 for =.05 and degrees of freedom = 2 (df = k-s-1 = 4-1-1 = 2), =5.99 • Hence distribution is uniform