I know what the answer to this is, but I haven;t got a clue how to arrive at it. Could someone please explain how I would approach this kind of problem?

Thanks.

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- Jul 3rd 2007, 03:36 AMGAdamsMedian
I know what the answer to this is, but I haven;t got a clue how to arrive at it. Could someone please explain how I would approach this kind of problem?

Thanks. - Jul 3rd 2007, 06:35 AMThePerfectHacker
Let $\displaystyle (x_1,x_2,...,x_n)$ $\displaystyle n$ numbers is increasing order (or non-decreasing). If $\displaystyle n$ is odd then $\displaystyle x_{\frac{n+1}{2}}$ is the location of the median. If even then $\displaystyle \mbox{Average}\left( x_{\frac{n}{2}}, x_{\frac{n+1}{2}} \right)$.

Since you have $\displaystyle n=100$ the answer is the average between the 50th and the 51st number when placed in increasing order. - Jul 3rd 2007, 11:58 AMCaptainBlack
The median is in the interval 10<=x<20m as there are 31 of 100 for x<10, and 55 of 100 for x<20, so the median is in the range 10<=x<20.

RonL