Difference between revisions of "1990 AJHSME Problems/Problem 20"
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==Solution== | ==Solution== | ||
| − | Let <math>S</math> be the sum of all the incomes but the largest one. For the actual data, the mean is <math>\frac{S+98000}{1000}</math>, and for the incorrect data the mean is <math>\frac{S+980000}{1000}</math>. The difference is | + | Let <math>S</math> be the sum of all the incomes but the largest one. For the actual data, the mean is <math>\frac{S+98000}{1000}</math>, and for the incorrect data the mean is <math>\frac{S+980000}{1000}</math>. The difference is <math>882\rightarrow \boxed{\text{The answer is A}}</math> |
==See Also== | ==See Also== | ||
Revision as of 20:54, 28 January 2020
Problem
The annual incomes of 
families range from 
dollars to 
dollars. In error, the largest income was entered on the computer as 
dollars. The difference between the mean of the incorrect data and the mean of the actual data is
Solution
Let 
be the sum of all the incomes but the largest one. For the actual data, the mean is 
, and for the incorrect data the mean is 
. The difference is 
See Also
| 1990 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
