Median

**GAdams** · July 3rd 2007, 03:36 AM

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.

**ThePerfectHacker** · July 3rd 2007, 06:35 AM

Let $(x_1,x_2,...,x_n)$ $n$ numbers is increasing order (or non-decreasing). If $n$ is odd then $x_{\frac{n+1}{2}}$ is the location of the median. If even then $\mbox{Average}\left( x_{\frac{n}{2}}, x_{\frac{n+1}{2}} \right)$ .

Since you have $n=100$ the answer is the average between the 50th and the 51st number when placed in increasing order.

**CaptainBlack** · July 3rd 2007, 11:58 AM

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

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