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Set X has 5 integers: a, b, c, d, and e. If m is the mean

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Set X has 5 integers: a, b, c, d, and e. If m is the mean [#permalink] New post 14 Jan 2012, 21:40
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Set X has 5 integers: a ,b ,c ,d , and e. If m is the mean and D, where D = sqrt{ [(a-m)^2+(b-m)^2+(c-m)^2+(d-m)^2+(e-m)^2]/5}, is the standard deviation of the set X, is D>2?

(1) a ,b ,c ,d , and e are different integers
(2) m is an integer not equal to any elements of the set
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Re: Standard Deviation and Mean [#permalink] New post 15 Jan 2012, 13:31
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Hi, there! I'm happy to help with this! :)

First, I'll say this is a very difficult question. If a question of this difficulty appears on the GMAT, it will be one of the harder questions, a question for which you probably will spend a tad more than the allotted 2 min/ question (if you are efficient on the easy questions, you will have a little more time for the more difficult questions like this).

Question: Set X has 5 integers: a ,b ,c ,d , and e. If m is the mean and D, where D = sqrt{ [(a-m)^2+(b-m)^2+(c-m)^2+(d-m)^2+(e-m)^2]/5}, is the standard deviation of the set X, is D>2?

Notice, this is the ordinary definition of standard deviation. If standard deviation is not a familiar topic, I would suggest reviewing it a bit in whatever review material you have.

Notice, also, the real crux of the question is: can we make the standard deviation small enough? There's no problem making a standard deviation way more than 2 -- for example, if the set is {100, 200, 300, 400, 1000}, then it satisfies both statements and has D >2. It's always easy to pick big numbers and numbers widely spaced out to increase the standard deviation. The trick is: can we make the numbers close enough together so that it's still possible that D is less than or equal to 2?

Statement #1: a ,b ,c ,d , and e are different integers
This eliminates the most obvious choice for a small standard deviation, viz, make all five numbers equal. If all the numbers of a set are equal, the standard deviation is zero, but that's not a possibility here. Well, if they have to be all different, the closest they can still be is consecutive, for example: X = {1, 2, 3, 4, 5}. Then, m = 3, and D = sqrt(((1-3)^2 + (2-3)^2 + (3-3)^2 + (4-3)^2 + (5-3)^2)/5) = sqrt((4 + 1 + 0 + 1 + 4)/5) = sqrt(10/5) = sqrt(2), which is less than 2. So, given statement #1, it's possible make a choice that has a D less than 2, and it's always possible to make other choices and make D huge, so there's no way to decide whether or not D > 2. Statement #1, by itself, is insufficient.

Statement #2: m is an integer not equal to any elements of the set
This is a tricky statement. It tells us (a) that the mean is an integer, not something true for a general set of five integers, and (b) this integer, the mean, is *not* equal to any of the five elements of the set. So, we can't have all five integers the same (since the mean would also be the same as them), and we can't have five consecutive integers. If we want a small standard deviation, we need the numbers still close together, so make some of them the same, with a kind of "hole" in the middle for the mean. For example, the set {1, 1, 1, 3, 3} does not have a integer mean, but the set {1, 1, 1, 3, 4} does have an integer mean: m = 2. (We were leaving the entry 2 vacated on the list, hoping the mean would land there.) So, this latter set, X = {1, 1, 1, 3, 4} satisfies statement #2, and has a standard deviation of D = sqrt(((1-2)^2 + (1-2)^2 + (1-2)^2 + (3-2)^2 + (4-2)^2)/5) =sqrt((1 + 1 + 1 + 1 + 4)/5) =sqrt(8/5), which is less than 2. Again, given statement #2, it's possible make a choice that has a D less than 2, and it's always possible to make other choices and make D huge, so there's no way to decide whether or not D > 2. Statement #2, by itself, is insufficient.

Combined Statements #! & #2:
This combination is particularly tricky. Neither of the sets we concocted under the separate statements will work here. We need five different integer, and the mean is going to be an integer not equal to the members of the set. This means, if we want a small standard deviation, we need to construct a set of integers close to each other, but with a "hole" in the middle where the mean will fall. If I try {1, 2, 3, 5, 6}, that has a mean of 3.4, a sum of 17. To have a mean that's an integer, the sum must be a multiple of 5. We are shooing for a sum of 20, which would be a mean of 4, which is why we are intentionally excluding 4 from the list. We could just add 3 the highest number to get {1, 2, 3, 5, 9}, but having one particularly huge number increases the standard deviation more than having a few sorta big numbers. Outliers enormously inflate the standard deviation. Starting with {1, 2, 3, 5, 6}, I can't add anything to the lower numbers without either duplicating a number in the set, or getting 4, which I trying to exclude. I can add +1 to 5 and +2 to 6, to get {1, 2, 3, 6, 8}. I believe that is as close as you can make a set of five integers with the constraints that (a) all are different, (b) the mean is an integer, and (c) the integer-mean is not a member of the set. With this set, X = {1, 2, 3, 6, 8}, the mean is m = 4, and the standard deviation is D = sqrt(((1-4)^2 + (2-4)^2 + (3-4)^2 + (6-4)^2 + (8-4)^2)/5) =sqrt((9 + 4 + 1 + 4 + 16)/5 = sqrt(34/5) > sqrt(6) > 2. In other words, we have constructed the tightest, most compact group of five integers permitted under the combined conditions, and even with our best choice, we are unable to make the standard deviation less than 2. The two conditions, combined, absolutely require that the standard deviation is more than 2. The combined statements are sufficient.

Answer = C

Does that make sense? Please let me know if you have any questions on what I've said here.

Mike :)
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Mike McGarry
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Re: Standard Deviation and Mean [#permalink] New post 15 Jan 2012, 13:49
Thanks Mike...i found this quite tough, not sure whether i can solve it in test quickly enough :(
Re: Standard Deviation and Mean   [#permalink] 15 Jan 2012, 13:49
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