Almost every website I find disagrees with you on this, and when I learned statistics, I learned that the word “average” had three common meanings.
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Edit: that said, my error was in suggesting that “expected value” could refer to median or mode. It looks, from a bit of Googling, that it only refers to the mean.
They are wrong.
Including mode makes them measure of central tendencies and people repeating misinformation still makes it wrong.
Let me ask you this: what is the definition of an average? I’m not asking for examples, but when you say the median is an average, what definition does it meet to be called that.
Definition of average.
An average is a function that maps a set of data to one and only one value (which is why mode is not an average) that:
Lies between the smallest and largest value
If a value smaller than the average value is added to the set and the average function is rerun, the average value either stays the same or decreases.
If a value larger than the average value is added to the set and the average function is rerun, the average value either stays the same or increases.
If the average value is added to the set and the average function is rerun, the average value stays the same.
If you look it up, most sites will say the definition of average is add up all of the numbers and divide by how many you have. That comes from the error that many people (including most elementary teachers) that average = mean = arithmetic mean.
So there are two ways to look at this.
The technically correct definition which excludes mode and allows for any function to be an average as long as it meets the definition - for example min and max are both technically averages but no where close to a central tendency.
Use it as in the common usage and is incorrect but since (almost) everyone uses it incorrectly no great loss, right?
Not exactly, it is a weighted mean. And to the point of language - WHICH mean? There are many.
Using the term “expected value” as a synonym for “mean” is IMO almost always wrong. For certain it’s wrong if we’re talking about multiple trials of something. Which in effect we are by saying the cop could be anywhere along the road.
Um, no? That’s not how words work? “Average” clearly has, as one of its accepted meanings, “mean, median, and mode.” You may not like that, but that doesn’t invalidate that meaning.
Can you offer an authoritative cite that specifically excludes “mode” from the meaning of “average”?
The expected value of a random variable only means one thing, namely an average of the outcomes weighted by probabilities, i.e., for a discrete variable \text{E}[X] = \sum_i x_i \mathop{\text{Pr}}[X=x_i].
And I would argue with the answer to number 2.
If the cop is standing at the unknown spot during the entire trip then you are really asking about the probability they are at a specific spot on the road and it is based on distance so the expected value is the arithmetic mean.
If it is a situation that a cop pulls up just as you are passing them, then the probability is based on your time matching theirs and the answer would be the one given by Gyrate.
Since the problem
I believe the answer is the arithmetic mean of 45mph. If you want the desired answer of 50mph the question would be something like
For a cop coming up on a side street just as you pass on the highway, what is the expected value of the car’s speed?
Every average is weighted, just sometimes the weights are uniform. 
The “average speed of the car” is a time-weighted average: [(60 mph)*(1 hr)+(30 mph)*(1 hr)] / (1 hr + 1hr ) = 45 mph
The “expected value of the cop’s measurement” is a position-weighted average: [(60 mph)*(60 mi) + (30 mph)*(30 mi)] / (60 mi + 30 mi) = 50 mph.
Divide the route into 90 equal segments, and the cop can be along any of them. At spots 1-60, the speed is 60 mph. At spots 61-90, the speed is 30 mph. Your data table has 60 values of 60, and 30 values of 30. The arithmetic mean of these values is 50.
Divide it into 90,000 equal segments, or 90 million equal segments, and a similar analysis applies.
How else would you figure it?
Of course, the fun of this problem is that English is imprecise. A precisely formulated question would be entirely in mathematical notation and be inaccessible to the general public. I tried to describe the problem in a way an average* person could understand.
Ultimately, there’s a difference between the time-averaged speed and the position-averaged speed. But how to explain that without a lot of math to a general audience?
*I see what I did there.
I like how you not only noticed what you did there, but you replied to yourself too.
Holy Self-Referential Writing, Batman!! Douglas Hofstadter would be so proud.
I’m about as average as they come.
And “expected value” to me means the cop will either see the car doing 60, or 30, and never 50 or any other speed, as this is the real world, and there were only two possibilities…
This reminds me of a much simpler problem that my father posed to me when I was a teenager. You have to assume instant acceleration and deceleration as needed (i.e no speeding up or slowing down, just instant transition to another speed).
Problem: a car travels one mile at 30 mph. How fast would it have to travel the second mile to average 60 mph over the two miles?
Answer: either it’s impossible, or theoretically at the speed of light. To drive one mile at 30 mpm takes two minutes. To drive two miles at 60 mph takes two minutes. So their (real world) time was up after the first mile.
I was pleased with myself at the time to have gotten it right. But it did take me a few minutes to think of the correct approach.
Totally not proper math terminology but if someone did not know that “expected value” is a math buzzterm, then you’re sorta not wrong in your own little world.
I found it interesting that that’s even the definition that Wolfram Mathworld (which is pretty mathematically sophisticated) gives.
In technical contexts, I’ve only ever seen the word “average” used to mean the mean. Median, mode, etc. are often called “measures of central tendency”, or “typical values”, or some such, but not averages.
In non-technical contexts, who knows what someone will use any given term to mean, because a lot of folks using those terms don’t know themselves.
Since this is MPSIMS and not FQ…
There is also a non-zero albeit very unlikely possibility that the answer to question 2 is NOTHING. Because the cop is dead.
You did say an unknown spot ON the road. If the observation time coincides with the car traversal time, an intersection of the two at either 60 or 30 miles per hour could result in a fatality!
Just to be complete in our observations.
You’re right. I don’t know why but I was thinking they were both 1 mile lengths when I wrote that.
Please Send Coffee.
I think that’s part of the “illusion” of the problem. When we think of traveling for two equal units, our brains think of the unit as length. It’s much trickier to think of the units as time. At least, it is for me.
But how mean are you?
This is my take as well. I wouldn’t consider the “expected value” to be a number that could literally not possibly be correct.