Sorry for the essay but its long .
Flooding has washed out one of the tracks of a railroad company. The railroad has two parallel tracks from baker to Cooper, but only on usable track from Cooper to Daniel, as shown below ( i drew it in the attachment ). Having only one usable track disrupts the usual schedule. Until it is repaired, the washed-out track will remain unusable. If the train leaving Baker arrives at Cooper first, it has to wait until the train leaving Daniel arrives at Cooper.
(look at picture)
Every day at noon a train leaves Baker heading for Daniel and another leaves Daniel heading for baker.
Assume that the length of time, X, it takes the train leaving Baker to get to Cooper is normally distributed with a mean of 170 minutes and a standard deviation of 20 minutes.
Assume that the length of time, Y, it takes the train leaving Daniel to get to Cooper is normally distributed with a mean of 200 minutes and a standard deviation of 10 minutes.
These two travel times are independent.
1.) What is the distribution of Y-X
2.) Over the long run, what proportion of the days will the train from Baker have to wait at Cooper for the train from Daniel to arrive?
3.) How long should the railroad company delay the departure of the train from Baker so that the probability that it has to wait is only 0.01?
2) Train from Baker has to wait at Cooper for the train from Daniel to arrive if X < Y => Y - X > 0 => U > 0 ....
3) Let the delay time be T minutes.
0.01 = Pr(X + T < Y) = Pr(Y - X > T) = Pr(U > T).
So find the value of a such that Pr(U > T) = 0.01.
Im sorry but i still dont know how to do number 2 at all. I think i see what your saying though, but not really.
and on number 3 i dont understand what the variables are in this, other than T which you told me :
0.01 = Pr(X + T < Y) = Pr(Y - X > T) = Pr(U > T).
Q2. U is defined in post #2 and its distribution given. Calculate Pr(U > 0).
Q3. U is defined in post #2. Calculate the value of T such that Pr(U > T) = 0.01. This is an inverse normal problem.
The real problem is that you lack sufficient understanding of the necessary background to answer this question. Take the advice I've given you - go back and thoroughly review the normal distribution and how to do calculations with it.
I have no idea where you got what you wrote? Are you a teacher? If you were how would you approach a student?
Im just trying to think of other ways to maybe open your mind to explain it to me because im lost.
we did normal distributions first 2 weeks of school and now what we are doing is getting back into them. I didnt forget about them, I just dont know how this applies to it.
I have told you in post #2 that U = Y - X follows a normal distribution with mean 30 and variance 500. That is the answer to Q1.
You're expected to be able to do basic calculations with a normal distribution. That is what Q2 and Q3 is asking you to do.
I have told you in Q2 to calculate Pr(U > 0). This is because you're need Pr(Y > X > 0).
I have told you in Q3 to calculate the value of T such that Pr(U > T) = 0.01. This is because you need Pr(X + T < Y).
Now go back and think about the definitions of X and Y given in the question.
I have shown you exactly what you need to do. Now it's up to you.
p.s. All my students understand the concept of reciprocal responsibility - I explain it very clearly to them in the very first class.
So i got .9101 doing normal cdf.
So thats saying that 91% of the days the train will have to wait at copperhead for the train from diamondback to arrive?
And you said in Q3 I user the INV normal. I have inv normal, but just like cdf, its been so long since ive used them i forgot what to put where.
Thanks for your help and patience with me, by the way.
No, I dont have the information, and my school is closed all week and i want to get this done before the weekend.
So, i remember my teacher setting up my calculator or showing me how to get it into help mode, so that like, for example, if i did normalcdf it would have showed me what i needed to enter.
If i can find how to activate that help guidance thing again i will be in good shape.