Challenging Probability Question for Math Experts?

Question

DETAILS
In a certain city, buses run every 30 minutes on average. At a particular station, buses are scheduled to arrive at 08:00, 08:30, etc. However, because of unpredictable traffic conditions, the actual arrival times are normally distributed about the scheduled arrival times with a standard deviation of 5 minutes. i.e. bus arrival times are N(08:00, 5^2), N(08:30, 5^2), etc.

INTERMEDIATE QUESTION
A passenger, trying to catch the bus, shows up at the station at exactly 08:00. What is his expected waiting time?

HARDER QUESTION
Suppose he shows up at 07:57 (3 minutes before the scheduled arrival time). What is now his expected waiting time?

CHALLENGE QUESTION
At what time must he show up in order to minimize his waiting time?
What would be the worst time to show up (which maximizes his waiting time)?

magicbright: you're on the right track.
You can find E(X|X>0) by working directly with the probability density function for the normal distribution.

http://en.wikipedia.org/wiki/Normal_distribution#Probability_density_function

Someone wanted clarification on this. If the bus arrives before 8, it does not wait until 8. Ignore any time delay between arrival and departure.

HINT:
Consider working directly with the probability density function for the normal distribution. And consider the role of the error function (erf) in performing the necessary integration.
http://en.wikipedia.org/wiki/Normal_distribution#Probability_density_function
http://en.wikipedia.org/wiki/Error_function

GENERAL SOLUTION

Passenger arrives T minutes before 08:00.
Let x = time the bus arrives, using 08:00 as a reference (μ = 0, σ = 5).
The probability density function for the normal distribution is
f(x) = 1/[σ √(2π)] * exp[-x2 / 2σ2]

Let W = waiting time = T + x
If he arrives before the bus, then he’ll wait T + x minutes. If he misses the 08:00 bus, he will have to wait for the 08:30 bus.
His expected waiting time,
E(W) = ∫{-T to inf} (T+x)*f(x)*dx + (T+30)*P(x<-T)

Evaluating each integral separately
∫{-T to inf} T*f(x)*dx = T/2 * [1 + erf (T/{σ√2}) ]
P(X < -T) = ∫{-inf to -T} f(x)*dx = 1/2 * [1 – erf (T/{σ√2}) ]
∫{-T to inf} x*f(x)*dx = σ/√(2π) * exp [-T2 / 2σ2]

Putting all terms together
E(W) = T + σ/√(2π) *exp [-T2 / 2σ2] + 15* [1 – erf (T/{σ√2}) ]

For T = 0, E(W) = 17.0 mins
For T = 3, E(W) = 12.9 mins

To find the min and max of E(W),
dE/dT = 1 – (T + 30) / [σ√(2π)] * exp [-T^2 / (2σ^2) ]
Setting dE/dT = 0 and solving numerically yields
E(W)_min = 10.2 mins when T = 7.4 mins
E(W)_max = 21.5 mins when T = -5.8 mins.

Ideal time = 07:53
Worst time = 08:06

Due to the nature of the problem, it was better to work with the probability density function for the normal distribution than the tables.
The intermediate problem involved integrating x*exp(-x2)
The harder problems involved integrating exp(-x^2). For this it is necessary to have some knowledge of the error function (erf). Matlab and Excel have the erf function built in.

Answer 1

Intermediate question.
For the waiting person, there are two scenarios: 1- the bus (X~ N(0,5) I am using 8:00 as the origin here) comes prior to 8:00 (P=0.5), which means he has to wait for the next bus to come. Starting at 8:00 the distribution of time till the next bus arrival (I’ll call it Y) is Normal(30, 5), assuming bus arrivals are independent from each other. So, if the first bus comes prior to 8:00, the expecting waiting time for the next one is equal to 0.5*30=15 min. The aggregate expected waiting time is then equal to 0.5*E(X|X>0)+15. Now we only have to calculate E(X|X>0) to obtain the answer. I don’t have an answer for that (just too much of math for me), I can only give you a formula that should calculate it: E(X|X>0)= the integral (from 0 to infinty of 2-2*Ф(t/5) dt).
The Harder question than is limited to changing the mu parameter of distributions of Y by 3 minutes (Y~N(33,5). X is still normal N(0,5)). Find the answer P(X<-3)*33+E(X| X>-3)*P(X>-3). As before, the whole problem comes to finding E(X|X>-3).

Answer 2

The solutions to the real/fake questions are the two real. all of the different questions are if truth be told the comparable, you in simple terms ought to divide the type of adventure that fulfill what you're searching for for via the type of all achieveable activities. Q1. There are 6 achieveable activities (you the two roll a a million,2, 3,4,5 or a 6). yet in ordinary terms one adventure is what you're searching for for, this is rolling a 5. So the respond is a million/6. Q2. returned, 6 achieveable adventure. 3 contain even numbers (2, 4 and 6) so answer is 3/6, which simplifies to a million/2. Q5. There are 12 (3+4+5) achieveable consequences, you ought to pick the 1st pink, 2nd pink, third pink, first blue, 2nd blue etc. right here, 4 activities are what you're searching for for, you pick the 1st blue, 2nd blue, third blue etc. So the respond is, 4/12 which simplifies to a million/3. Q6. returned there are 12 (3+6+3) achieveable consequences, 3 of which we are looking for for. So the respond is 3/12 which simplifies to a million/4.

Answer 3

Can anyone teach me how to do this after the best asnwer is chosen? I need help with probability and statistics.