1. The arithmetic mean of a data set is 46 and the standard deviation of the set is 4. Which value is exactly 1.5 standard deviations from the arithmetic mean of the set?
2. for how many integers n is 2^n=n^2作者: beyondgmatqin 时间: 2013-3-13 13:51
这道题的意思应该是 哪一个数是恰好距离平均数1.5个standard deviation？
所以46+(1.5x4)=52 或46-(1.5X4)=40
A. 42 B.44.5 C.47.5 D. 50 E. 52
答案有52 选E作者: beyondgmatqin 时间: 2013-3-13 13:53
=> 2^n = n^2
Taking nth root on both sides
=> 2 = (n^2)^1/n
=> 2 = n ^ 2/n
Lets consider positive even multiples of 2 for n (since LHS = 2)
For n = 2
=> 2 = 2 ^ 2/2 - First value that satisfier
For n = 4
=> 2 = 4 ^ 2/4 - Second value that satisfier
For n = 8
=> 2 = 8 ^ 2/8 - Doesnt satisfy
For n = 16
=> 2 = 16 ^ 2/16 - Doesnt satisfy
Two values. Ans = C作者: beyondgmatqin 时间: 2013-3-13 13:53
希望对你你有帮助哈。。作者: howie0420 时间: 2013-3-13 19:58
楼主，巧用google：
1.http://www.manhattangmat.com/for ... viation-t13232.html
take the standard deviation given (4) multiplied by the number of standard deviations the problem is interested in (1.5). In this case we get 6. Add and subtract that to the mean of 46; we get -1.5 standard deviations at 40 and +1.5 standard deviations at 52.
2.http://www.manhattangmat.com/for ... -2-n-n-2-t5030.html
2 and 4.
hopefully, 2 jumped out at you pretty fast; if n = 2, then the two sides of the equation are 2^2 and 2^2. i'm pretty sure that those are going to be equal.
you'll discover n = 4 through raw experimentation; there isn't any better way, unfortunately.
if you want to be confident that these are the only solutions, you have to watch the behavior of 2^n and n^2 as you get further and further away from 4. the pattern you'll observe is that 2^n begins to grow much, much faster than does n^2, making it clear that the two expressions won't be equal for any larger values.
the equality is definitely impossible for negative integers, because 2^(negative integer) is a fraction, while (negative integer)^2 is not. therefore, you don't have to worry about negative integers.作者: cclovelife 时间: 2013-3-16 07:56
Terrific! Thanks~
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