Calculating your average

[fourdown00]

how do you calculate your average? like when people say their average of 5 how do they get that?
5 solves in a row?
pick 5 solves you have done?

thanks for the help guys

 

[hdskull]

5 consecutive solves with the best and worst times taken out indicated by parenthesis. e.g.

times: 22, 18, (24), 20, (15)
then the avg would be: (22+18+20)/3 = 20.

 

[Johannes91]

times: 22, 18, (24), 20, (15)
then the avg would be: (22+18+20)/3 = 20.

You should add 0.5.

 

[Jason]

why add 0.5?

 

[Johannes91]

http://tinyurl.com/2dpd7t
http://www.speedcubing.com/records/records-rules.html

 

[Jason]

that site says "If you are measuring your time with an instrument that does not record the tenths of a second then you must add 0.5 seconds to your average time calculated above".
how do you know the fellow isn't using a stackmat or whatever?

 

[hdskull]

sorry, assume they were exact solves (22.00, etc) timed on a stackmat

 

[Joël]

sorry, assume they were exact solves (22.00, etc) timed on a stackmat

Yeah, that's what I though.

Nobody uses timers that time in seconds these days. (And if they do, they don't post it).

 

[Johannes91]

how do you know the fellow isn't using a stackmat or whatever?

I just assumed so because the numbers were integers.

 

[Radu]

i do the average out of 12

 

[AvGalen]

You can use this formula:

(Sum(Results) - Min(Results) - Max(Results)) /
(Count(Results) - Count(Min(Results)) - Count(Max(Results)))

(and yes, Count(Min(Results)) and Count(Max(Results)) could be simplified to 1)

That would mean (based on 22.00 + 18.00 + 24.00 + 20.00 + 15.00)
(99.00 - 15.00 - 24.00) / (5 - 1 - 1)
60.00 / 3 = 20.00

 

[adragast]

By the way, in competition, I can only see average of 5... should we still continue to speak about average of 12 ? All PC-timers I have seen are making averages of 12... Should we on the contrary have average of 12 in competition (for the finals of world championships for example) ?

 

[masterofthebass]

I think an average of 12 is too much for any competition. People usually don't do 12 solves great in a row, plus it's way too time consuming. Even if people can solve it sub-15, thats 15*12 + scrambling time that a competitor has to partake in. I mean thats 3 minutes of timer time for the 3x3. Now imagine that for a 5x5...

 

[Kristoffer]

times: 22, 18, (24), 20, (15)
then the avg would be: (22+18+20)/3 = 20.

You should add 0.5.

Well i guess he used easy numbers just to show how it works..:rolleyes:

 

[TheBB]

You can use this formula:

(Sum(Results) - Min(Results) - Max(Results)) /
(Count(Results) - Count(Min(Results)) - Count(Max(Results)))

Well, if you wanted a cumbersome way to express a stunningly simple concept, yes.

 

[hdskull]

or Arnaud's explanation in simpler terms:

(all times added - best time - worst time)/(number of solves - 2)

 

[Johannes91]

Or, for those who speak J:

(+/%#)_1}.1}.(/:y){y=. 22.00 18.00 24.00 20.00 15.00