Averaging Minutes and Seconds

Date: 07/03/2002 at 01:59:09
From: Corey Tharp
Subject: Averaging Minutes and Seconds

I need the average time of 20 different times.  If I punch it in on a 
calculator, it is incorrect due to time being on a 60 unit cycle 
compared to a 10 unit cycle on the calculator.  I understand this, but
still can't figure out how to get around it.  

I read a problem in your archives that was similar, but I still don't 
get it.  Can you please help?  Any feedback would be greatly 
appreciated.

Date: 07/03/2002 at 14:09:30
From: Doctor Rick
Subject: Re: Averaging Minutes and Seconds

Hi, Corey.

I think you're saying that you punched in, say, 4.25 for 4 minutes 
and 25 seconds, and it didn't come out right. The correct way is 
first to convert seconds to minutes: for instance, 25 seconds = 25/60 
minutes, so 4 minutes, 25 seconds = 4 + 25/60 = 4.1667 minutes. 
Convert each time from minutes and seconds to decimal minutes in this 
way, then average.

Perhaps we can speed things up. First add up the seconds ONLY, for 
all 20 times. Then divide by 60 to convert all the seconds to 
minutes. Then continue adding the minutes part of each time. Finally, 
divide by the number of times (DON'T count the minutes and seconds 
separately) to do the averaging.

Does this make sense to you? Here's why it works. When you average 3 
times, you should do this:

  (2m,25s + 3m,10s + 5m,50s)/3

We need to express all in minutes so we can add:

  (2+25/60 + 3+10/60 + 5+50/60)/3

We can rearrange the terms using the commutative property of addition:

  (2 + 3 + 5 + 25/60 + 10/60 + 50/60)/3

We can combine the divisions into a single division using the 
distributive property of multiplication by 1/60 (which is what 
division by 60 is):

  (2 + 3 + 5 + (25 + 10 + 50)/60)/3

That's what I did, only I put the seconds first so we wouldn't need 
parentheses with the calculator:

  25 + 10 + 50 =     85
  / 60 =              1.41667
  + 2 + 3 + 5 =      11.41667
  / 3 =               3.80556 minutes

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/