Here’s a little physics puzzle to while away some time.
A car takes a trip along a straight road. For the first hour, it travels at 60 mph. Without turning, it travels at 30 mph for the next hour.
1. What is the average speed of the car?
2. For a cop standing at an unknown spot on the road, what is the expected value of the car’s speed?
There’s no tricks here, it’s just simple math. Hint: the answers to the two questions is different.
The way i see it :-
car does 90 miles in 2 hours. Therefore average speed is 45mph
Don’t understand the question !
I guess the cop would expect the car to be driving at or within the posted speed limit.
What speed can a cop expect to see the car passing them by at?
never mind
The car will have covered twice as much distance in the first hour as it did in the second. So for a cop at a random spot there’s a 2/3 chance the car will be going 60.
Expected value - I’m winging the calculation here, but is it (60 x 2) + (30 x 1) = 150, and 150/3 = 50mph? Or is this nonsense?
That is exactly right.
I concur with @pjd for the first question, and @gyrate for the second, assuming that “expected speed” is referring to a mean, not a mode or median.
I would think the median and mode would also be the same, given a reasonable reading of the problem, which I take to be “for a cop likely to be standing at any point along the route with equal probability”.
Though, sure, I would say it is a mild weakness in the problem statement that it is necessary for the reader to make such an assumption rather than have it explicitly stated.
I don’t think that’s right.
There is an arbitrary number of spaces that the cop could be standing on the road. On 2/3 of them, the driver is going at 60 mph, and at 1/3, the driver is going at 30 mph. If you list each of those spaces as a separate instance, the most common speed value, the mode, is 60 mph. If you list them in order from lowest to highest, the middle value, the median, is also 60 mph.
Sorry, that was unclear. I meant I expect the median and mode to be the same. You’re right the mean is different.
Oh, gotcha–I see! Yeah, we agree.
Yay! I mathed right!
What does ‘expected value’ mean here? How is he measuring the speed?
Expected value is another way to say the average value.
Doesn’t matter how the speed is measured, assume perfect enough.
If the cop times the value from the start to the end then 45MPH is the expected value. Otherwise it’s anywhere between the maximum reverse speed and the maximum forward speed the car is capable of travelling.
ETA: You did specify the operating speed of the car for each mile so unless the cop measures a time span that crosses the midpoint then it’s either 30MPH or 60MPH. If the time span measured crosses the midpoint it’s somewhere between 30MPH and 60MPH. The question also assumes the car can immediately deaccelerate from 60MPH to 30MPH at the midpoint.
Well, the average speed was asked for in part 1 and it’s 45 mph.
That’s the average speed of the car over the entire trip. The average speed of the car as it passes the cop is different.
[pedant] No, the expected value of the speed of the car as it passes the cop is different.[/pedant]
Yep, there’s lots of quibbles that can be made. Feel free to do the math for whatever assumptions you’d like to use.
Nope. Weighted average. That’s why it is not equal to the mean.
E(x) = Σ[xi X p(xi)]
Also, it is an error to say the average. There are many averages
arithmetic mean
geometric mean
quadratic mean
Heronean mean
median
midrange
etc.
I’ll also point out for those discussing the mode - mode is not an average. Not sure if people were using it in that way.