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Topic: MEAN, MEDIAN, NEITHER (Read 817 times) |
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kapiwu
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MEAN, MEDIAN, NEITHER
« on: Dec 12th, 2007, 9:05pm » |
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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 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.
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Grimbal
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Re: MEAN, MEDIAN, NEITHER
« Reply #1 on: Dec 13th, 2007, 4:30am » |
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It is the median among the cars that you met in an overtake. Any car that you overtook (or that you overtook once more than you overtook it) started in front of you and ended behind you. So its speed is lower than yours. Inversely, every car that overtook you goes faster than you. That is why you are in the middle of the pack, and that is the median. How much faster or slower the other cars go you cannot tell, so it doesn't have to be the average speed.
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SMQ
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Re: MEAN, MEDIAN, NEITHER
« Reply #2 on: Dec 13th, 2007, 6:56am » |
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I'm going to go with the mean; consider the following distribution:
1 car per mile traveling 79 mph
2 cars per mile traveling 75 mph
3 cars per mile traveling 73 mph
4 cars per mile traveling 63 mph
and 1 car (Alice) per mile traveling x mph
To find x such that the number of cars she passed = the number of cars which passed her, note that for each speed s, the number of cars at speed s which Alice passed per hour is d(s) * (x - s). For example, if her speed (x) is 4 mph faster then s, and there is 1 car per mile traveling at speed s, it will take her 1/4 hour to cover the mile between each of those cars, so she will pass 4 of them each hour. Now, we want to find x such that the net number of cars she passed is 0, that is: sum over i of d(si) * (x - si) = 0. So sum over i of d(si) * x - sum over i of d(si) * si = 0, and x = (sum over i of d(si) * si) / (sum over i of d(si)). But this is just (sum of the speeds of all cars) / (number of cars), which is the mean speed.
So, in the distribution above, Alice's speed is 70 mph, and she was passed by and passed 28 cars per hour. Note also that her speed is neither the median speed (which is 73 mph) not even the mode (which is 63 mph).
--SMQ
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--SMQ
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Grimbal
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Re: MEAN, MEDIAN, NEITHER
« Reply #3 on: Dec 13th, 2007, 8:26am » |
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Hm... while I still stand by my answer that it is the median of the car Alice met, it appears that it is not what the question asked. It was about the cars on the road. Alice meets more of the extremely fast or extremely slow cars, so the cars she met is not representative of the cars on the road.
I agree with SMQ's analysis.
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