Question

Every morning Mary randomly decides on one of three possible ways to get to work. She makes her choice so that all three choices are equally likely. The three choices are described as follows: • Choice A (Drives the highway): The highway has no traffic lights but has the possibility of accidents. The number of accidents on the highway for the hour preceding Mary’s trip, X, follows a Poisson distribution with an average of 2. The time (minutes) it takes her to get to work is affected by the number of accidents in the hour preceding her trip due to clean up. The time (in minutes) it takes her is given by T = 54.5 + 5X. • Choice B (Drives through town): Suppose there is no possibility of being slowed down by accidents while going through town. However, going through town she must pass through 10 traffic lights. Suppose all traffic lights act independently from one another and for each there is a probability of 0.5 that she will have to stop and wait (because it is red). Let Y be the number of lights she will stop and wait at. The time (in minutes) it takes her is given by T = 58.5 + Y. • Choice C (Takes the train): Trains arrive for pick-up every 5 minutes. If the train has room, it will take her exactly 50 minutes to get to work. If an arriving train is full she will have to wait an additional 5 minutes until the next train arrives. Trains going through the station will arrive full with probability 0.75, and thus she cannot get on and will have to wait until the next train. Suppose it takes Mary exactly 5 minutes to get to the train station and she always arrives at the station just as a train arrives. Let Z be the number of trains she’ll see until she can finally board (the train isn’t full). The time (in minutes) it takes her is given by T = 50 + 5Z.

a) Which choice should she make every morning to minimize her expected travel time?

b) On one morning Mary starts her journey to work at 7am. Suppose it is necessary that she is at work at or before 8:00 am. Which route should she take to maximize the probability that she is at work at or before 8:00am?

Answer 1

We will initially find the expectation of each random variable

A) E(X)=2 as X follows poisson distribution with an average 2minutes

E(T)=54.5+5*E(X)=64.5

B) Let Y₁, Y₂,.....,Y₁₀ be 10 bernouli random variables taking two values 0 and 1, 0 denotes waiting and 1 denotes not waiting at traffic signal

Since all the random variables are independant the sum follows binomial distribution with mean (np) ie E(Y)=E(Y₁+ Y₂+.....+Y₁₀)=10*0.5=5 minutes

E(T)=58.5+E(Y)=63.5

C) Z is a random variable following geometric distribution ie number of trains she should see before could board a train which is not full will follow geometric distribution with probability of success p=0.25 and q=1-p=0.75(Probability that Trains going through the station will arrive full)

E(Z)= 1/p=1/0.25=4

E(T)=50+5*E(Z)=70

a) She should choose B to minimize her expected travelling time

b)If Choice is A

  

  

  

If the choice is B

If the choice is C

so the choice which makes her arrival to work before 8 am is C since it gives the maximum probability to be there before 8

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