| Discussion: | Research Area |
| Topic: | Simple arithmetic problem "99%" can't solve |
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| Subject: | Simple arithmetic problem "99%" can't solve |
| Author: | Sonny |
| Date: | Jun 15 2004 |
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statistician, F. J. Moroney and solved it incorrectly. Repeatedly, in years
since, in university classes, none solved it correctly. I've the original ONLINE
in http://members.fortunecity.com/jonhays/stndtsk1.htm
(along with other problems I required of my students, but many teachers and
professors could not solve). Another version: in auto trip, of 4 spans (starting
in town traffic, to out-of-town), 1st span averages 10mph; 2nd span, 20 mph;
3rd, 30mph; 4th, 40mph. What average speed represents these 4 average speeds?
Many persons think arithmetic mean (AM) is "the average". (Moroney said
"average" is "representative".) AM = 1/4(10 + 20 + 30 + 40)mph = 1/4(100)mph =
25mph. This incurs 30%+ relative error. The average speed is distance traversed
divided by time traversing. We need
a representivve distance. The least common multiple of (10, 20, 30, 40)mi. is
120mi. Time: at 10mph, 12 hrs; at 20mph, 6 hours; at 30mph, 4 hours; at 40mph,
30. Total: (12 + 6 + 4 + 3)hours = 25hours. Then we have 480mi/25hr = 19.2mph.
Relative error:(25-19.2)/19.2% = 5,8/19.2%, approx. 30%. But a different
"average", the harmonic mean (HM), yields exactly the correct answer. Easily
calculated as AM of reciprocals of values. 1/HM = 1/4(1/10 + 1/20 + 1/30 + 1/40)
=
1/4(12 + 6+ 4 + 3)/120 = (1/4)(25/120) = 25/480 = 1/HM, or HM = 480/25 = 19.2.
(Ancient Pythagoreans used HMs to construct our chromatic musical scale.) Why
isn't this more generallly taught?
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