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 = 4.41667 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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.