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 Tags averages , statistics question

 26th November 2008, 03:29 PM #1 Tatsu New Blood   Join Date: Mar 2004 Posts: 24 A question about statistics and averages My company tracks telephone performance by the number of calls abandoned, i.e. the percentage of callers that hang up before we answer their call. Our goal is <=5%. Every month I receive a report that gives the percentage of dropped calls for the previous month, along with the raw data. Here are the numbers for last month: R = Received: 1437 A = Answered: 1339 a = Abandoned: 76 The number of abandoned calls is not a simple subtraction of Answered from Received but instead only unanswered calls that end after a certain time threshold. That way we are not penalized for calls that end before we have the chance to answer. If I were calculating our performance I would do it like this: a/R=0.05288 or 5.3% If I were being completely merciless: a/(R-A)=0.0681 or 6.8% Here is how it is actually being calculated: a/{R-[(R-A)-a]}=0.0537 or 5.4% So my question is: What reason could they have for doing it that way? Please have mercy, I am not a mathematician.
 26th November 2008, 03:49 PM #2 Tanja Comfortably Numb     Join Date: Nov 2003 Posts: 3,001 I presume the important thing is what the remaining 22 calls were. If I understood correctly, they were calls that you did not have any chance to answer, so they may be viewed as, say, "invalid calls", so you don't want to include them in the total number of calls, so they are taken out of the equation alltogether.
 26th November 2008, 03:54 PM #3 GodMark2 Graduate Poster     Join Date: Oct 2005 Location: Oregon, USA Posts: 1,041 Originally Posted by Tatsu a/{R-[(R-A)-a]}=0.0537 or 5.4% So my question is: What reason could they have for doing it that way? Please have mercy, I am not a mathematician. Originally Posted by Tanja I presume the important thing is what the remaining 22 calls were. If I understood correctly, they were calls that you did not have any chance to answer, so they may be viewed as, say, "invalid calls", so you don't want to include them in the total number of calls, so they are taken out of the equation alltogether. Or more simply, the ratio of abandoned to the total of abandoned and answered, but not counting the impossible-to-answer, which the number received would count. a/(A+a) Which is what that final equation simplifies to. The total received is unimportant. __________________ Knowing that we do not know, it does not necessarily follow that we can not know.
 26th November 2008, 07:26 PM #4 Tatsu New Blood   Join Date: Mar 2004 Posts: 24 Ahh... when you put it that way it makes more sense. Many thanks to you Tanja and GodMark2!
 27th November 2008, 04:27 PM #5 DazzaD Critical Thinker   Join Date: Jul 2006 Location: Romford Posts: 303 Originally Posted by Tatsu Here is how it is actually being calculated: a/{R-[(R-A)-a]}=0.0537 or 5.4% As a matter of interest... is that actually how they quote the formula? I.e. has someone said or printed that that is how they do it? Exactly as written above?
 27th November 2008, 10:53 PM #6 Tatsu New Blood   Join Date: Mar 2004 Posts: 24 No. It is embarrassing to admit, but that is how I worked it out for myself. I realized that I had gone the long way around the barn as soon as I read GodMark2's reply.

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