20 Golf PlayersDate: 04/11/97 at 12:01:25 From: Mark Mullen Subject: Variance in average team handicaps There are twenty players with golf handicap between 6 and 32. (6,6,8,9,10,11,13,15,15,16,18,19,21,23,26,27,29,30,31,32). The problem is to divide the twenty players into four teams of five players each, with the variance in the average team handicaps being no more than one or two. Is there a mathematical formula that can be used to do this rather than using trial and error, as is done every spring? Date: 04/11/97 at 15:48:52 From: Doctor Steven Subject: Re: Variance in average team handicaps This simple method should get a close enough approximation to the minimum variance of the means. Start by placing the golfers in order, listing them down one column and then going up on the next column and so on and so on. For your golfing problem we get this arrangement (with the first row being the first team, etc, etc): 6 15 15 27 29; Average Handicap: 18.4 6 13 16 26 30; Average Handicap: 18.2 8 11 18 23 31; Average Handicap: 18.2 9 10 19 21 32; Average Handicap: 18.2 Variance of Averages: .01 Standard Deviation of Averages: .1 This method should land fairly close to the optimal strategy (and it's probably a lot less work!). Hope this helps. -Doctor Steven, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/