Math 462 / 562 Final Exam Fall 2012

Name: ______This is a closed book exam. You may use a calculator and the formulas handed out with the exam. Show all work and explain any reasoning which is not clear from the computations. Turn in this exam along with your answers. However, don't write your answers on the exam itself; leave them on the pages with your work. Also turn in the formulas; put them on the formula pile.

1. A company uses a special optical sensor in its production line. Unfortunately the sensor wears out rather rapidly. In fact the company has observed that the lifetime S of the sensor follows a uniform distribution between 1 and 10 days. More precisely the probability that the sensor will fail sometime during the kth day after it is installed is 1/10 for k=1,2,…, 10 and the probability is 0 for any other value of k. So Pr{S = k} = 1/10 for k=1,2,…, 10.

The sensor costs $100 to replace. However, when the sensor fails during production, the production line has to be shut down for the rest of the day at an additional cost of $400. This is in addition to the $100 that it costs to replace the sensor. However, this additional cost is not incurred if the sensor is replaced in the evening when the production line is not being used. To do this requires replacing a sensor that has not failed.

Suppose the company replaces the sensor after it has been used n days if it has not been replaced sooner because of failure. Let T be the time between replacements. So T = S if S £ n and T = n if S > n.

a. (6 points) What is the probability the sensor will fail before it is replaced? Your answer will involve n. For example, if n = 3 then the probability the sensor will fail before it is replaced is 3/10.

b. (6 points) What is the expected cost, E(C), of a sensor including the additional cost if it fails before it is replaced? Your answer will involve n.

c. (6 points) What is the expected time, E(T), between replacements? You can use the fact that 1+2+3+…+m=m(m+1)/2. Your answer will involve n. For example, if n = 3 then the only values T can have are 1, 2 and 3.

d. (7 points) Find n to minimize the average daily cost of a sensor.

2. You work for a market research firm and you are doing a telephone survey about consumer attitudes toward buying a new car. You estimate that for any given person you call there is a 30% chance that they will participate in the survey and whether any person agrees to participate is independent of whether all the other people agree to participate. Calculate the numerical value for your answer to each of the following.

a. (12 points) What is the probability that exactly 3 of 7 people you call agree to participate in the survey?

b. (13 points) What is the probability that at least 3 of 7 people you call agree to participate in the survey?

3. A cell phone company models the time T that it takes for a typical phone call by a random variable with probability density function f(t) = . Time is measured in minutes.

a. (8 points) Find the probability that a phone call lasts between 1 and 2 minutes.

b. (8 points) Find the cumulative distribution function F(t) = Pr{T £ t} of T.

c. (9 points) Find the mean of T.

4. Sam is playing his favorite computer game during a thunderstorm. The time between lightning strikes on a power line is exponentially distributed with mean equal to 30 minutes. Assume that each time lightning strikes a power line the power is interrupted and the computer crashes and the game is ruined. The time it takes Sam to play his game is exponentially distributed with mean equal to 45 minutes. Assume the times between lightning strikes and the times to play games are all independent of each other. Answer the following. Give actual numbers for the answers.

a. (12 points) Find the probability that there is no more than 2 lightning strikes in the next 3 hours.

b. (13 points) What is the probability that Sam will be able to play a game uninterrupted?


Solutions

1. a. Pr{sensor fails before it is replaced} = Pr{S £ n} = Pr{S = 1} + Pr{S = 2} + … + Pr{S = n} = n/10.

b. E(C) = 100 + 400Pr{S £ n} = 100 + 400n/10 = 100 + 40n = 20(5 + 2n).

c. E(T) = (1)Pr{T = 1} + (2)Pr{T = 2} + … + (n – 1)Pr{T = n - 1} + (n)Pr{T = n} = (1)(1/10)+(2)(1/10)+ … + (n – 1)(1/10) + n(1/10 + 1 – n/10) = (1/10)(1 + 2 + … + n + 10n – n2) = (1/10)(n(n + 1)/2 + 10n – n2) = (1/20)(n2 + n + 20n – 2n2) = (1/20)(21n – n2).

d. z = E(C)/E(T) = 400(5 + 2n)/(21n – n2). dz/dn = 400[(21n – n2)(2) – (5 + 2n)(21 – 2n)]/(21n – n2)2 = 400[42n – 2n2 – 105 + 10n – 42n + 4n2]/(21n – n2)2 = 400(2n2 + 10n -105)/(21n – n2)2. So dz/dn = 0 Û 2n2 + 10n - 105 = 0 Û n = (- 10 ± )/4 = (- 5 ± )/2 »= (- 5 ± 15.33)/2 = 5.16 and 10.16. We are not interested in the negative value, so n = 5.15. However, n must be an integer. If n = 5 then E(C)/E(T) = 400(15)/(5)(16) = 75. If n = 6 then E(C)/E(T) = 400(17)/(6)(15) » 75.6. So n=5 is optimal.

2. a. Let N = number that participate. This is a binomial probability. Pr{N = 3} = (0.3)3(0.7)4 = (35)(0.027)(0.2401) » 0.2269

b. Pr{N ³ 3} = 1 - Pr{N = 0, 1 or 2}. Note that Pr{N = 0, 1 or 2} is the sum of binomial probabilities. Pr{N=0,1or2} = (0.7)7 + (0.3)1(0.7)6 + (0.3)2(0.7)5 = 0.0824 – 0.2471 – 0.3176 = 0.6471. So Pr{N³3}= 1 – 0.6471 = 0.3529.

3. a. Pr{1 < T < 2} = = = = - = - =

b. F(t) = Pr{T £ t} = . If t < 0 then F(t) = 0. If t > 4 then F(t) = 1. If 0 £ t £ 4 then F(t)= = = = =

c. E{T} = = = = = 4 - =

4. Let's measure time in hours. Let S = time to play a game and T = time between lightning strikes. Then S is exponential with density function f(s)= and T is exponential with density function g(t)=.

a. Let N(t) be the number of lightning strikes in a time interval of length t. In class we saw that Pr{N(t) = n} = where l is the parameter in the associated exponential distribution. In this case l = 2, so Pr{N(t) = n} = We want Pr{N(3) = 0, 1 or 2} = Pr{N(3)=0} + Pr{N(3)=1} + Pr{N(3)=2}. From the above Pr{N(3) = n} = . Therefore Pr{N(3) = 0, 1 or 2} = + + = (1 + 6 + 18)e-6 = 25e-6 » (25)(0.00248) » 0.06197.

b. We need to find Pr{S < T}. One has Pr{S < T} = f(s)g(t) dsdt where A = {(s,t): s < t}. Since f(s) = 0 for s0 and g(t) = 0 for t < 0 one has Pr{S < T} = where A+ = {(s,t): 0 < s < t}. So Pr{ST}= = = = = .