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Listing all the possibilities would take time. Would there be a faster and more efficent way to find the median?

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    There are 40 cars with 25mpg. There are 30+10=40 cars with 28 or more mpg. Those cancel each other, leaving just two columns. – vadim123 Aug 12 '14 at 02:09

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The median of a data set is the "middle" point. Let's do this by area of the "bars" on the graph. The first bar has an area of 40. The second of 50, third of 40, fourth of 30 and last of 10. If we "shade" the first bar, we must shade a total area of 40 on the right. So the 28 and 29 bars have a combined area of 40 and therefore we shade them. Now we shade the smaller of the two remaining, which is the 27 bar of area 40. Now we shade the bottom area of the 26 bar and see that we still have an area of 10 in that bar and nothing else to cancel with. Therefore, the only possibility for the median is 26.

This is basically the same as listing them all out, except we are canceling groups of them at a time, rather than individually.

BeaumontTaz
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  • How is the bar with nothing to cancel out with the median? –  Aug 12 '14 at 02:16
  • @Ben, let's relate this to you list. If we listed them all out, then at the point when we have an area of 10 left we have everything cross off except this: $\ldots,26,26,26,26,26,26,26,26,26,26,\ldots$ and we know this is the middle 10 numbers from our large list. What is the median of this list? – BeaumontTaz Aug 12 '14 at 02:17
  • how do u know to make 40 the gold standard that 28mpg and 29mpg are to combine to form? –  Aug 12 '14 at 02:21
  • @Ben, Because the area of $10$ that is $29$mpg and the area of $30$ that is $28$mpg have the same combined area ($10+30=40$) as the area of $40$ that is $25$mpg. We really skipped a step on that one. We should have crossed off the area of $10$ from $29$mpg and the same area of $10$ in the $25$mpg bar. And then cross of an area of $30$ from the $28$mpg bar and the same area in the $25$mpg bar. That would result in the same step that we saw when we grouped the last two together. – BeaumontTaz Aug 12 '14 at 02:22
  • I don't understand how your method of canceling out the bars works. I think I would understand better with smaller numbers. Can your method be applied to solve the median for a simpler set of numbers such as $"3,4,5, and 7?"$ –  Aug 12 '14 at 02:24
  • The method would that the you have an area of $1$ above the number $3$, an area of $1$ above the $4$ and so on. Each have an area of $1$. The area just represents how many of each of them we have. A better example that works better to explain the method would be this data set: $(1,1,2,3,3)$. Where $1$'s have an area of $2$, $2$'s an area of $1$ and $3$'s an area of 2. We cross off the same area on the bottom as we do that top; therefore we cross of the area of $2$ from the $1$ and the area of $2$ from the $3$. It's the same as crossing off the bottom 2 and the top 2 numbers simultaneously. – BeaumontTaz Aug 12 '14 at 02:27
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    @Ben, the other way to solve that problem would be to cross off the bottom term and the top term. Leaving us with $(1,2,3)$ and then doing it again. But if we know we can afford to cross off more than one on each side, why not make out work easier and cross them both off at the same time. That's all we're trying to employ here. – BeaumontTaz Aug 12 '14 at 02:31
  • I don't see how adding the bars in your method is equivalent to crossing out "Both at the same time". Can you explain? –  Aug 12 '14 at 18:46
  • Ok. I see how adding the bars is equivalent. –  Aug 12 '14 at 18:49