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Topic: HARD: Mean, Median, Neither (Read 5471 times) |
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tim
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HARD: Mean, Median, Neither
« on: Jul 26th, 2002, 3:18am » |
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The answer is the mean. With a little maths, this is fairly easy to see:
Let Alice's speed be 'A', and use the conventions in the notes to the problem, i.e. there is a uniform density of d(s) cars per mile moving at speed s, all integers.
Then in the one hour, Alice travels A miles. The cars travelling at speed s cover s miles, so Alice passes (A-s) miles worth of such cars. Since there are d(s) cars per mile, Alice passes d(s)(A-s) of them. If s>A, then the result is negative, signifying that the cars passed Alice. We sum over all s to get the net number of cars passed.
We know that the net number of cars passed is zero, i.e. as many cars passed Alice as Alice passed other cars. Hence
Sum(d(s)(A-s)) = 0,
A Sum(d(s)) = Sum(s d(s)), which means
A = Sum(s d(s))/Sum(d(s)).
The term on the right hand side is the definition of the mean, so Bob was right the first time.
This problem was greatly simplifed by having integers for all quantities. In the more general case, Alice's speed would be equal to the mean in the limit of infinite time spent on the road, but need not be equal for any finite time.
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| « Last Edit: Jul 27th, 2002, 1:01am by william wu » |
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Kozo Morimoto
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Re: Mean, Median, Neither (solved)
« Reply #1 on: Jul 26th, 2002, 5:26am » |
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It has to be median.
Say Alice was travelling at 2M/h
Alice passed the car travelling at 1M/h
Alice was passed by a car travelling at 2997M/h
Mean speed of the 3 cars = 1000M/h which Alice is nowhere near. But Alice's speed of 2M/h IS the median by definition.
Extending this example to n:
speed of cars Alice passed (slower than Alice) are S1,S2,S3 ... Sn
speed of cars Alice was passed by (faster then Alice) are F1,F2,F3 ... Fn
then if you ranked all the cars by speed all the Sx will be before Alice and all the Fx will be after Alice, therefore Alice's speed will be the median, but not the mean.
This all assumes that all cars (including Alice's) are travelling in the same direction. She could have 'passed' cars coming from the opposite direction.
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Joe Marshall
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Re: Mean, Median, Neither (solved)
« Reply #2 on: Jul 26th, 2002, 5:37am » |
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Actually, the speed of Alice's car need not be the mean or the median. Kozo Morimoto's reply proves that Alice's speed is the median speed of the following set of cars: Alice's car, plus all the cars she passed, plus all the cars which passed her. However, if there are other cars on the road which neither pass nor are passed by Alice (e.g. a car travelling faster than Alice, which starts ahead of her), Alice's speed will not be the median of _all_ the cars on the road.
This may not be in the spirit of the problem, however.
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Rhaokarr
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Re: Mean, Median, Neither (solved)
« Reply #3 on: Jul 26th, 2002, 9:27pm » |
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I didn't understand what this problem was about - I thought the definition of median was the centermost event. Since the number of events greater equalled the number of events lesser, Alice's car speed would have to be the central event, therefore the median by definition.
Is the question missing something? I don't think it's a trick question, since you could keep adding data to make any answer wrong (eg what about cars that started in front of her that she never saw, blah blah blah)
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tim
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Re: Mean, Median, Neither (solved)
« Reply #4 on: Jul 27th, 2002, 1:08am » |
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I didn't have any trouble interpreting the problem, I just used its statement as clarified in the notes. The density of cars of a given speed (all integer-valued) is specified to be integer-valued and uniform throughout the road, repeating every mile.
This messes up Kozo's median argument -- if there is the minimum of one car per mile of road travelling at 2997 mi/hr, then she is passed by 2995 of them, not just one. If the number of cars travelling at 1 mi/hr is the same, then Alice only passes one of them. In order to pass as many 1 mi/hr cars as she is passed by 2997 mi/hr cars, there has to be 2995 times as many slower cars on the road.
The correct answer really is the mean, as proved by following the mathematics in my first post.
The argument extends equally well to cars travelling in the opposite direction -- you can consider them to have negative speed and just plug them into the formula like everything else.
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Kozo Morimoto
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Re: HARD: Mean, Median, Neither
« Reply #5 on: Jul 27th, 2002, 3:31am » |
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"In order to pass as many 1 mi/hr cars as she is passed by 2997 mi/hr cars, there has to be 2995 times as many slower cars on the road."
Why can't this be true? d(s) does not have to be linear - it could be totally random by using a lookup table. d(1) could be 2,000,000 and d(13) could be 96 you don't know.
I guess I'm not understanding the note. Can you explain how many 5mi/h cars pass Alice IF Alice is going at 2mi/h and d(5)=7? Is it (5-2) * 7 = 21?
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tim
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Re: HARD: Mean, Median, Neither
« Reply #6 on: Jul 27th, 2002, 4:03am » |
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Yes, there is no problem with there being 2995 cars per mile doing 1 mi/hr and 1 car per mile doing 2997 mi/hr. But in that case, Alice isn't doing the median speed any more (which is 1 mi/hr) 
She is doing the mean speed though: (2995*1 + 1*2997)/2996 = 2.
Regarding the second part, yes you understand it correctly. In the hour, Alice travels 2 miles. All the cars doing 5 mi/hr travel 5 miles, and so 3 miles worth of cars pass Alice. That's 21 cars, as you calculated.
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Kozo Morimoto
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Re: HARD: Mean, Median, Neither
« Reply #7 on: Jul 27th, 2002, 5:26am » |
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"Yes, there is no problem with there being 2995 cars per mile doing 1 mi/hr and 1 car per mile doing 2997 mi/hr. But in that case, Alice isn't doing the median speed any more (which is 1 mi/hr)
She is doing the mean speed though: (2995*1 + 1*2997)/2996 = 2."
No, she gets passed by equal number as the cars she passes so the formula is:
(2995*1 + 2995*2997+2)/(2995+2995+1) which I don't think is 2.
So in the example:
Alice travels at 3mi/h
d(1)=7
d(10)=2
So she passes 14 cars at 1mi/h and she gets passed by 14 cars doing 10mi/h
Alice's 3mi/h is the median
mean = (14cars * 1mi/h + 14cars * 10mi/h + 3mi/h)/(14 + 14 + 1) = 157/29 = 5.413 <> 3
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Alex Harris
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Re: HARD: Mean, Median, Neither
« Reply #8 on: Jul 27th, 2002, 6:16am » |
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I'm with tim on this. Remember that the mean we're talking about is the mean of the cars on the road, not the mean of the cars she passed. So the calculation for mean is actually:
(7 cars * 1 mph + 2 cars * 10 mph ) / 9 cars = 3 mph.
I left out Alice's car in the calculation since it doesn't affect the question of whether she is travelling the mean speed or not.
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Kozo Morimoto
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Re: HARD: Mean, Median, Neither
« Reply #9 on: Jul 27th, 2002, 6:42am » |
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How does she know how many cars are there on the road unless she passes it or is passed by it.
Using your logic how can there be only 2 cars at 10mi/h in the example? If she was passed by 14 of them, there gotta be AT LEAST 14 if not more (the cars on the road that she didn't see) There can't be less cars on the road than she saw - that doesn't make any sense.
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Alex Harris
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Re: HARD: Mean, Median, Neither
« Reply #10 on: Jul 27th, 2002, 7:07am » |
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You seem to be confused about the question.
"Note: Assume that any car on the road drives at a constant nonzero speed of s miles per hour, where s is a positive inte- ger. And suppose that for each s, the cars driving at speed s are spaced uniformly, with d(s) cars per mile, d(s) being an integer. And because each mile looks the same as any other by the uniformity hypothesis, we can take mean and median to refer to the set of cars in a fixed one-mile segment, the half-open interval [M, M+1), at some instant."
There are 2 10-mph cars and 7 1-mph cars per 1 mile segment.
Remember
"Bob: your speed must have been the mean of the speeds of the cars on the road. "
Mean/median of the cars on the road is not the same thing as the mean/median of the cars which passed her. If the question was regarding the cars that passed her then her speed would indeed be the median, but the problem would be completely trivial.
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-D-
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Re: HARD: Mean, Median, Neither
« Reply #11 on: Jul 27th, 2002, 12:10pm » |
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Sorry Folks, Tim is definately right.
I think that you'd really have to consider an infinite amount of time to prove the answer, because there are situations that can be found when analyzing finite time blocks that would attempt to disprove the argument. But what you really get is that for a finite amount of time, it's possible to have a speed "region" for which the observation holds true. And the speed region centers around the mean.
I derived Tims answer by trying to think about it in simple terms. Maybe this will help.
The cars that are passing her are passing her at certain rate. And she is passing other cars are a certain rate. What you really want to find is what is Alice's speed such that the rate of cars passing her is the same as the rate she is passing cars.
So. here goes. To find out the rate cars pass her, you find the relative speed of the car vs Alice.
A = Alice's speed
s = signifies one speed "group"
(A - s) = relative speed of "speed group" in MPH.
d(s) = density of cars in terms of CARS/MILE.
=> d(s)(A-s) = is in units of CARS/HOUR
This is a transformation of relative velocity to number of cars passed in a particular speed group. They are essentially the same thing.
Now rule that ties this all together is: The number of cars passed is the same as the number of cars passing her. Reinterpret to the sum of all cars passed or passing is zero.
** denote a summation as E
and the boundries (not needed) are
n = -oo to n = oo (infinity)
Ed(s)(A-s) = 0
.. the rest of the math was proven above.
-D-
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Alex Harris
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Re: HARD: Mean, Median, Neither
« Reply #12 on: Jul 27th, 2002, 1:27pm » |
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We don't need to mess about with infinities. Any hour that Alice drives looks exactly like the first. Tim's first post is a proof IMO. He could elaborate more on justifications but the essentials are all there and correct.
Technically I should add the statement "when there are other cars on the road" to my earlier statement about not needing to count Alice's speed in the formula. In the trivial case of no other cars on the road we want the mean to be defined by Alice.
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Bob MyKnob
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Re: HARD: Mean, Median, Neither
« Reply #13 on: Jul 28th, 2002, 5:15pm » |
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Couldn't Alice be going at 0 mph? I mean that's only if an equal number of cars drive on both sides, since to her it would look like she's passing the other cars thanks to relativity. It's probably not the right answer, but it went through my head so I though I would throw it out.
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bartleby
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Re: HARD: Mean, Median, Neither
« Reply #14 on: Jul 29th, 2002, 10:48am » |
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What if the other cars on the road were not travelling constant speeds?
What if cars were speeding up and slowing down?
What if a car kept speeding up and slowing down, passing and getting passed by Alice repeatedly?
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-D-
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Re: HARD: Mean, Median, Neither
« Reply #15 on: Jul 29th, 2002, 10:54am » |
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bartelby: the riddle provides that all cars travel at a constant speed s and at every speed there is a "density" of d(s) cars/mile. Those situations would not occur.
Bob: the solution that Tim presented allows for cars to be traveling in the opposite direction. s becomes negative and therefore the relative speed of that vehicle is A- (-s).
-D-
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william wu
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Re: HARD: Mean, Median, Neither
« Reply #16 on: Aug 8th, 2002, 9:06pm » |
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Some e-mails I got which argue that the answer is "neither":
E-MAIL A
Note: Assume that any car on the road drives at a constant nonzero speed of s miles per hour, where s is a positive inte- ger. And suppose that for each s, the cars driving at speed s are spaced uniformly, with
d(s) cars per mile, d(s) being an integer. And because each mile looks the same as any other by the uniformity hypothesis, we can take mean and median to refer to the set of cars in a fixed one-mile segment, the half-open interval [M, M+1), at some instant.
This is easy to prove with a simple example. Suppose there are 100 parked cars per mile (speed == 0), and 1 car per mile traveling at 100 mph, and Alice was going 1 mph. In one hour, she'll pass 100 cars and 100 cars will have passed her. Yet the median speed is clearly 0, and the mean speed is (100*1+0*100)/101 = 100/101 != 1, not counting Alice, and (100*1+0*100+1*1)/102 = 101/102 != 1, counting Alice. So Alice's speed cannot be said to be either the mean or the median, although it is certainly possible to construct scenarios where it is.
E-MAIL B
Alice was more likely travelling at the mean speed of the traffic, but need not have been (especially since she has a small data set). To visualize this, imagine the traffic as a collection of rolling marbles, all of equal mass. As the marbles roll along, the only way for your marble to be passed by as many as pass you is to have the number of marbles behind stay constant and the number of marbles ahead stay constant.
A special case of this is where your marble is at the center of mass (assuming equal-mass marbles), and remains there. Remaining at the center of masss of a collection of same-mass marbles means travelling at the average speed of those marbles.
For the puzzle's wording, using cars, you get the equivalent if you consider a group of cars evenly split front and back. If the ones behind are passing at the same rate as the ones ahead are being passed, then Alice is moving with the "center of mass" of the group of cars. (We couldn't be sure of this if you hadn't specified that the cars all drive at constant speed.)
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tim
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Re: HARD: Mean, Median, Neither
« Reply #17 on: Aug 8th, 2002, 11:34pm » |
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Email A is incorrect: In that scenario Alice is passed by 99 cars, not 100. Since the number of cars she passes does not match the number of cars who pass her, of course she is not doing the mean speed.
Email B is considering the more general problem of arbitrary distributions of cars, at a guess. The problem is that in the more general case, the "mean" is not well defined. It is certainly possible to construct scenarios where Alice is traveller at either vastly greater or less speed than either of the mean or median.
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Eric Yeh
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Re: HARD: Mean, Median, Neither
« Reply #18 on: Aug 9th, 2002, 6:04am » |
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I just saw this puzzle and thread (due to its recent charge to the front of the forum). Tim is completely correct, it is indeed the mean. His explanation of the mistake in A is spot on, and regarding B, I may have read it too quickly, but I didn't see any explanation of why it didn't have to be the mean, as claimed in the first sentence. I think it may be a mistake based on the "small sample size", which is actually precise in this case with the integer values of time and speed.
Best,
Eric
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AlexH
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Re: HARD: Mean, Median, Neither
« Reply #19 on: Aug 9th, 2002, 9:38am » |
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This really doesn't deserve a sticky 
Email A is wrong as indicated and Email B is actually supporting the mean even if it doesn't say so. Take some integer length road section --- Alice is travelling with the speed of the center of mass of the cars on that section. For any integer length time the length of road that has cars that pass or are passed by Alice is symmetric about her and intergral in length. She is travelling at the center of mass of those cars.
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Eric Yeh
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Re: HARD: Mean, Median, Neither
« Reply #20 on: Aug 9th, 2002, 10:47am » |
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I agree about the sticky. I think this one is pretty much resolved.
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| « Last Edit: Aug 9th, 2002, 10:48am by Eric Yeh » |
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James Fingas
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Re: HARD: Mean, Median, Neither
« Reply #21 on: Aug 29th, 2002, 11:31am » |
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Neither.
Alice was driving at 99 miles an hour, there were two cars almost one mile ahead when she started, going 98 and 97 miles an hour. She got off the highway after 49 miles.
No cars passed her. She passed no cars.
The mean and median are both 98 miles per hour.
All this shows is that we have to consider her going an infinite number of miles for the problem to have a unique answer.
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AlexH
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Re: HARD: Mean, Median, Neither
« Reply #22 on: Aug 29th, 2002, 12:41pm » |
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Check the problem description James. Your example violates the problems stated assumptions.
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James Fingas
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Re: HARD: Mean, Median, Neither
« Reply #23 on: Aug 29th, 2002, 1:16pm » |
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Thanks.
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