Mean or Median or Neither

[BMAD]

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.

Edited December 4, 2013 by BMAD

[bonanova]

If it's just one mile's worth of cars, no conclusion can be reached because it depends not only on her speed but her position in the group. If she was in front she could pass no one; if in back, no one could pass her.

But if OP means to join enough of those groups together that there is no practical limit of cars in front of her or behind her, then the answer is median. As many cars were faster (and would pass her) as were slower (and would be passed by her.

Edit:

I guess you need the added condition that the speeds were uniformly distributed.

If not, you could have cars driving at the speed of sound contributing disproportionately to the passing cars, or a set of parked cars contributing disproportionately to the cars passed.

With that condition, e.g. consecutive-integer speeds, median is the answer.

Without it, the answer is not certain.

I think.

Maybe even with consecutive-integer speeds, the car with median speed would pass more cars than pass her.)

[bonanova]

Three conveyor belts, a, b and c, move respectively at linear speeds of 1, 2 and 3 units per second.
Alice stands on belt b.

Case 1:
Markers are placed 1 unit apart on belts a and c.
At time t=0 Alice observes markers that go by her on the other belts.
After one minute Alice has passed 60 markers on belt a and has been passed by 60 markers on belt c.

Case 2:
Markers are placed 2 units apart on belt a and 2 markers per unit on belt c. The run is repeated.
After one minute Alice has passed 30 markers on belt a and has been passed by 120 markers on belt c.

Case 3:
Markers are placed 1 unit apart on belt a and 2 units apart on belt c. The speed of c is increased to 4 units/second. The run is repeated.
After one minute Alice has passed 60 markers on belt a and has been passed by 60 markers on belt c.

Case 4:
A fourth belt d is added. The speeds are respectively 1, 2, 3 and 4 units/second.
Markers are placed 1 unit apart on belt a and 3 units apart on belts c and d. The run is repeated.
After one minute Alice has passed 60 markers on belt a and has been passed by 20 markers on belt c and 40 more on belt d: a total of 60 markers.

Case 1: Alice moves at both mean and median speed. Passed / Passed By markers are equal in number.
Case 2: Alice moves at both mean and median speed. Passed / Passed By markers are unequal in number.
Case 3: Alice moves at median (but not mean) speed. Passed / Passed By markers are equal in number.
Case 4: Alice moves at neither mean nor median speed. Passed / Passed By markers are equal in number.

Having equal numbers of markers passing Alice and being passed by Alice places no constraint whatever on Alice's speed relative to the group.

[plasmid]

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.

Edited December 6, 2013 by plasmid

[bonanova]

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.

[gavinksong]

Alice was moving at the mean speed.

Suppose {a_1, ... a_n} are the speeds of all the cars in one mile that are faster than Alice,

{b_1, ... b_m} are the speeds of all the cars in one mile that are slower than Alice,

and C is Alice's speed.

Then the number of cars that Alice passes is (C - b_1) + (C - b_2) + ... (C - b_m),

and the number of cars that pass Alice is (a_1 - C) + (a_2 - C) + ... (a_n - C).

These equal each other, so we may infer...

(C - b_1) + (C - b_2) + ... (C - b_m) + (C - a_1) + (C - a_2) + ... (C - a_n) = 0

(n + m) * C = b_1 + b_2 + ... b_m + a_1 + a_2 + ... a_m

C = (b_1 + b_2 + ... b_m + a_1 + a_2 + ... a_m) / (n+m)

Thus, C is the average speed.

[gavinksong]

Three conveyor belts, a, b and c, move respectively at linear speeds of 1, 2 and 3 units per second.

Alice stands on belt b.

Case 1:

Markers are placed 1 unit apart on belts a and c.

At time t=0 Alice observes markers that go by her on the other belts.

After one minute Alice has passed 60 markers on belt a and has been passed by 60 markers on belt c.

Case 2:

Markers are placed 2 units apart on belt a and 2 markers per unit on belt c. The run is repeated.

After one minute Alice has passed 30 markers on belt a and has been passed by 120 markers on belt c.

Case 3:

Markers are placed 1 unit apart on belt a and 2 units apart on belt c. The speed of c is increased to 4 units/second. The run is repeated.

After one minute Alice has passed 60 markers on belt a and has been passed by 60 markers on belt c.

Case 4:

A fourth belt d is added. The speeds are respectively 1, 2, 3 and 4 units/second.

Markers are placed 1 unit apart on belt a and 3 units apart on belts c and d. The run is repeated.

After one minute Alice has passed 60 markers on belt a and has been passed by 20 markers on belt c and 40 more on belt d: a total of 60 markers.

Case 1: Alice moves at both mean and median speed. Passed / Passed By markers are equal in number.

Case 2: Alice moves at both mean and median speed. Passed / Passed By markers are unequal in number.

Case 3: Alice moves at median (but not mean) speed. Passed / Passed By markers are equal in number.

Case 4: Alice moves at neither mean nor median speed. Passed / Passed By markers are equal in number.

Having equal numbers of markers passing Alice and being passed by Alice places no constraint whatever on Alice's speed relative to the group.

In case 2, Alice is moving at

neither mean or median speed. Since there are half as many markers on conveyor belt a and twice as many markers on conveyor belt c, this changes the median and mean speed. The mean speed is now 13/5 > 2, and the median speed is 3.

In case 3, Alice is moving at the mean speed, not the median speed. Since there are half as many markers on belt c, the speed of belt c is now weighted by half its original weight when calculating the average. Since the speed of belt c was doubled, the average did not change.

In case 4... I think you get my point.

[bonanova]

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.