**Edited by BMAD, 04 December 2013 - 02:23 AM.**

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# Mean or Median or Neither

Started by BMAD, Dec 04 2013 02:22 AM

7 replies to this topic

### #1

Posted 04 December 2013 - 02:22 AM

Alice-- i was driving on a highway recently for one hour at a constant and very special speed.

Bob-- what was so special about it?

Alice-- the number of cars i passed was the same as the number of cars that passed me!

Bob-- your speed must have been the mean of the speeds of the cars on the road.

Alice-- or was it the median?

Bob-- these two are often confused. maybe it's neither? we'll have to think about this.

Was Alice's speed the mean, median, or neither?

Note: Assume that any car on the road drives at a constant nonzero speed of s miles per hour, where s is a positive integer. 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.

### #2

Posted 05 December 2013 - 08:48 AM

Spoiler for Looks like

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #3

Posted 06 December 2013 - 01:49 PM

Spoiler for On further review

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #4

Posted 06 December 2013 - 04:05 PM

In case 2 I think Alice is not traveling at the median speed, if you count the median of the markers rather than the median of the conveyor belts.

Edit: But this starts to touch on an important point. The way the OP is phrased, for each car there are an infinite number of identical 1-mile segments that repeat as if on a conveyor belt. A car gets counted for each time it's placed within a 1-mile segment, which is not necessarily the same as the number of times Alice sees one of those cars passing or getting passed.

Suppose there are cars moving at 45 MPH, 50 MPH, 55 MPH, and c/2 where c = speed of light. Neglecting relitivistic effects, even though there's only 1 warpspeeder per 1-mile segment, Alice will be passed by a ton of repeating 1-mile segments if she's going between 50 and 55.

Edit: But this starts to touch on an important point. The way the OP is phrased, for each car there are an infinite number of identical 1-mile segments that repeat as if on a conveyor belt. A car gets counted for each time it's placed within a 1-mile segment, which is not necessarily the same as the number of times Alice sees one of those cars passing or getting passed.

Suppose there are cars moving at 45 MPH, 50 MPH, 55 MPH, and c/2 where c = speed of light. Neglecting relitivistic effects, even though there's only 1 warpspeeder per 1-mile segment, Alice will be passed by a ton of repeating 1-mile segments if she's going between 50 and 55.

**Edited by plasmid, 06 December 2013 - 04:15 PM.**

### #5

Posted 07 December 2013 - 05:09 AM

Point is you can change cars passed or passing by adjusting their density independent of Alice's speed. She just can't be the slowest or fastest car.

At any speed other than slowest or fastest the passed and passing cars can be equal in number.

At any speed other than slowest or fastest the passed and passing cars can be equal in number.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #6

Posted 11 December 2013 - 05:45 AM

Spoiler for My answer...

### #7

Posted 11 December 2013 - 05:55 AM

Spoiler for On further review

Spoiler for There is a mistake...

### #8

Posted 11 December 2013 - 11:27 AM

It may not be clear whether mean (median) applies to the speeds of the lanes or the ensemble of cars. My assumption was the former; as I think of it now, the latter makes a better puzzle.

Edit: and it is what the OP says.

Edit: and it is what the OP says.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

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