Course 3 #20

[Gracewang]

Do you think there could be an error in the preliminary answer keys?
I still think the right answer for #20 is D instead of C.

[Macroman]

Do you think there could be an error in the preliminary answer keys?
I still think the right answer for #20 is D instead of C.

I too answered D on the exam. I ran a PC simulation and I am convinced that C is the correct answer. The trick is that the express rider can have a very long wait for an express train (heavy tail).

[Gracewang]

I think the expected waiting time for express rider is 6m and for "you" is 1.5m. I just don't understand why not?

[Macroman]

1. the waiting time is not what the problem asks for.

2. I did it where the local rider (me I guess) arrives at departure time so has no wait.

3. The express rider has an average wait of 12 min (if they arrive at a departure time) or 13.5 min if they don't know departure time).

4. I did a binomial expansion of probabilities and arrival time differences for the cases, they summed up to show that each rider has same arrival time.

ergo
p(express) dt = express-local
.25 0
.75(.25) -9
.75^2(.25) -6
.75^3(.25) -3
.75^4(.25) 0
.75^5(.25) +3
and so on

I took this out to 100 trains where the probability of the 1st express falling on that iteration was on the order of 10^(-11) and the difference of arrival time was nil...

There is an exam appropriate solution which is mentioned in another thread.

[Gracewang]

Here is what I thought:

The express rider has an average wait of 6 min( if they arrive at a departure time, it will be 12 min, but he won't be always so unlucky) till he gets an express train, so the total expected time for him is 6+16=22 min.

The local rider has an average wait of 1.5 min till he gets a local train with probability 0.75, an express train with probability 0.25, so the total expected time for him is 1.5 + 0.75*28 + 0.25* 16 = 26.5 min.

What's wrong with this? I couldn't figure out.

[3rookie]

Grace

This is what I posted earlier:

The express rider takes 28 minutes (12 min wait + 16 min ride)

The other rider waits an average of 1/20 hours (3 minutes) for a train.
.75 prob that it's local
.25 prob that it's express
3 minute wait + .75(28 minutes for local) + .25(16 min for express)=3 + 21 + 4 = 28 minutes
On average, they arrive at the same time

Hope this helps

[Gracewang]

Thank you. 3rookie.

Don't you think the expected waiting time for the express rider is 12/2 =6 min?
I think, from one express train departure to the next express arrival, the expected waiting time is 12 min. But the rider could arrive the station at any time instead of the second of departure. So he has an average waiting time of 6 min.
And same for the local rider.

[Macroman]

I'm interested to see the SOA's solution to this one. My binomial expansion isn't really valid because train arrival is Poisson.

I think Grace is right about expected arrival of the trains. I'm not convinced that the difference in expected arrival times of the trains translates directly into an answer.

It seems unlikely but not impossible that the SOA answer and most popular choice would both be incorrect.

[Mr. BoH]

Don't you think the expected waiting time for the express rider is 12/2 =6 min?
I think, from one express train departure to the next express arrival, the expected waiting time is 12 min. But the rider could arrive the station at any time instead of the second of departure. So he has an average waiting time of 6 min.
And same for the local rider.

No, his expected waiting time is 12 minutes, no matter when he arrives. Remember, Poisson is a memoryless process, so his expected waiting time is the same whether he missed the last train by 5 seconds or no train has gone by for an hour.

Your solution would only be appropriate if the problem stated "trains arrive at a regular interval of every 12 mintutes."

[Gracewang]

You are right. Mr. BoH.
Thank you very much.