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GMAT Statistics http://www.800score.com/forum/viewtopic.php?f=3&t=43 
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Author:  questioner [ Wed Aug 04, 2010 10:09 am ] 
Post subject:  GMAT Statistics 
J is a collection of four odd integers whose range is 4. The standard deviation of J must be one of how many numbers? A. 3 B. 4 C. 5 D. 6 E. 7 (B) For the sake of simplicity, let the smallest number in J be 1. Since the range is 4, the largest number in J is 5. You may recall that the standard deviation of a set of numbers is the square root of its variance. How many different variances are possible? { 1, 1, 1, 5 } and { 1, 5, 5, 5 } will have the same variance. Why? Remember that the variance is the average squared difference between each element and the mean of the set of numbers. The first set above has a mean of 2, whereas the mean of the second set is 4. In the first set, three elements are 1 unit from the mean and the other element is 3 units from the mean. This is true of the second set, too. These two sets, then, have a variance of 12/4. Similarly, { 1, 1, 3, 5 } (mean 2.5) and { 1, 3, 5, 5 } (mean 3.5) will have the same variance: in each set, two elements are 1.5 units from the mean, one element is 0.5 unit from the mean, and the remaining element is 2.5 units from the mean. These two sets have a variance of 11/4. Finally, {1, 3, 3, 5} and {1, 1, 5, 5} have variances of 8/4 and 16/4 respectively. So, 4 different variances and thus four different standard deviations are possible. Answer: B  How do you calculate the number of different sets possible? 
Author:  Gennadiy [ Wed Aug 04, 2010 1:18 pm ] 
Post subject:  Re: math (test 5, question 35): sets. 
The general form of such set is {n, m, k, n + 4}, where n  odd integer, m, k  odd integers such that n ≤ m ≤ k ≤ n + 4. For the sake of simplicity we assume that n = 1. We can do that because standard deviation remains the same if we add or deduct any fixed number to/from all the set members. So set has following form: {1, m, k, 5 }, m, k  odd integers such that 1 ≤ m ≤ k ≤ 5. So to calculate all the possible sets we plug in m and k that satisfy above mentioned criteria: {1, 1, 1, 5} {1, 1, 3, 5} {1, 1, 5, 5} {1, 3, 3, 5} {1, 3, 5, 5} {1, 5, 5, 5} 
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