Percentage calculation advice
Category: Science > Math
Asked by: drpangloss-ga
List Price: $15.00
05 Oct 2004 10:17 PDT
Expires: 04 Nov 2004 09:17 PST
Question ID: 410650
Sorry to say that it's been a while since my math and statistics courses and I'm trying to avoid any erroneous concept ("lying with statistics") errors. I am trying to calculate some percentages (no rounding applied in the example below) from two sets of tallies. I know pretty well how to handle (a.) and (b.). The (c.) percentage is over 100%. The (d.) percentage is a negative value. The (m.) percentage expresses the sum of the (a.) through (d.) values. a. 120 is 83% of 144 b. 195 is 93% of 208 c. 182 is 189% of 96 d. -91 is -124% of 73 m. 406 is 77% of 521 My question is about the effect of the exception (c.) or (d.) values upon (m.). [I actually have about sixty values almost all of which are like (a.) and (b.).] Specifically my first question is: what is the term that should be used to express the (m.) 77% value. Can it be termed the arithmetic mean percentage of the (a.) through (d.) values, or should it be called something else? Second question: is the (d.) calculation correct and/or does its [and (c.)s for that matter] inclusion in the set of numbers create a situation where the (m.) percentage value could be criticized. In other words, are any mathematical incongruities [probably nonrelevant but, for example, a sin such as dividing by zero] being introduced into the (m.) percentage that someone could take issue with?
Re: Percentage calculation advice
Answered By: mathtalk-ga on 05 Oct 2004 11:55 PDT
Hi, drpangloss-ga: Although the percentage found in (m) is not a "straight" arithmetic mean of the percentages found in (a) through (d), it is a "weighted" average (or weighted arithmetic mean) in which the percentages (a) through (d) are weighted according to their corresponding "bases": 144*(83%)+208*(93%)+96*(189%)+73*(-124%) 77% = ------------------------------------------ 144 + 208 + 96 + 73 I could quibble about your statement that you did not apply rounding, since the rounding rule you seem to have used is "truncation toward zero". But I checked your arithmetic, esp. in (c),(d), and (m), and found everything was at least consistently rounded by that rule. Along the lines of mathematical improprieties, I'd beware of using negative weights in a calculation like this. But it seems that all your weights (denominators of the intermediate calculations) will be positive, so I think it's okay to describe (m) as a weighted mean. regards, mathtalk-ga
rated this answer:
and gave an additional tip of:
Thanks for addressing my question perfectly and providing a most complete answer. I especially appreciate that you additionally caught my sloppy use of the term "no rounding" and provided appropriate terminology!
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