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^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
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.
