A friend at work mentioned that his daughter had been given an assignment in
maths to find a series of five numbers such that the median is less than the
mode, and the mode is equal to the mean. Naturally, I couldn't let this rest
until I'd worked it out, and eventually I got to proving that it's impossible.
Here's my proof - if someone finds a hole in it please let me know :-)

Let the five numbers be a, b, c, d, e, in ascending order. For there to be a mode that is not the median, two numbers have to be the same and every other number is different - those two numbers have to be a and b or d and e. Let's consider the case where they're d and e - the other case is symmetric. For the mean to equal the mode:

- (a+b+c+d+e)/5 = d. In other words:
- a+b+c+d+e = 5d. Since d = e,
- a+b+c = 5d - 2d. Therefore:
- (a+b+c)/3 = d.

Hopefully the next person who gets given this rather bizarre question will find this and get the answer without straining their brain coming up with cases. It is, of course, quite possible that the question had been garbled in between the teacher and me - it is, of course, trivial to think of a five number series where the median is less than the mean which in turn is less than the mode. Ah well, that's that off my brain now... :-)

Last updated: | path: personal | permanent link to this entry

All posts licensed under the CC-BY-NC license. Author Paul Wayper.

Main index
/ tbfw/
- © 2004-2016
Paul Wayper

Valid HTML5