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Is the standard deviation of numbers w,x,y and z greater than 2? 1) w=3 2) The average of the four numbers is 8

You might be able to see intuitively why the answer should be C here - if the average of our four-element set is 8, and one of the elements is equal to 3, then at least one of our elements is pretty far from the mean. That's certainly going to guarantee something about our standard deviation - the standard deviation can't be all that small here.

You can prove that the answer is C, at least if you remember how you find the standard deviation (not something you ever need to calculate on the GMAT, incidentally) - you find the distance from each element to the mean, square those distances, average these squares, then finally take the square root. If our mean is 8, and one of our elements is 3, then one of our distances to the mean is 5. So when we add the squares of the distances, our sum must be at least 5^2 = 25. We now take the average of this sum (so here divide by 4); this average is at least 25/4 = 6.25, and the square root of that is greater than 2.

So if we use both statements, the standard deviation must be greater than 2, and the answer is C. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Re: Is the standard deviation of numbers w, x, y and z greater than 2? [#permalink]

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10 Oct 2010, 16:17

1

This post received KUDOS

great clue by IanStewart

C

for the set to have a mean of 8 and smallest element to be 3 => sum of other 3 elements = 29 = 8*4 - 3

=> set can be 3, 9, 10, 10 or can have any other 3 number whose sum is 29. whichever those number are, the sum of the difference between those numbers and mean is always 5.

Is the standard deviation of numbers w, x, y and z greater than 2? [#permalink]

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10 Jan 2016, 23:19

Bunuel wrote:

cfpenteado wrote:

I would say the following:

St1) w=3

If all were equal, then SD, would be: w=3, x=3, y=3, z=3 SD=0

If all were different: w=3, x=0, y=-3, z=-6 SD>2

Insufficient

St2) Avg of w,x,y,z = 8

If all were equal, then SD, would be: w=2, x=2, y=2, z=2 SD=0

If all were different: w=-4, x=0, y=0, z=12 SD>2

Insufficient

St1 and St2)

If all were equal, then SD, would be: w=3, x=4, y=-3, z=4 SD=0

If all were different: w=3, x=-4, y=-3, z=12 SD>2

Insufficient

In your examples the average is never 8. Plus SD of 3, 4, -3, 4 is ~3, not 0.

St1) w=3

If all were equal, then SD, would be: w=3, x=3, y=3, z=3 SD=0

If all were different: w=3, x=0, y=-3, z=-6 SD>2

Insufficient

St2) Avg of w,x,y,z = 8

If all were equal, then SD, would be: w=8, x=8, y=8, z=8 SD=0

If all were different: w=-4, x=0, y=0, z=36 SD>2

Insufficient

St1 and St2)

w=3 so the rest will have to sum up to 29. I think by any way I try the set will always have a SD > 2 (more scattered data)

an example: w=3, x=5, y=12, z=12 avg = 8 SD > 2 always. In my opinion, SD always > 2. so C I think. Also, corrected my second statement. Thanks Bunuel Sorry Guys, I rushed on this one.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is the standard deviation of numbers w, x, y and z greater than 2?

(1) w = 3 (2) The average of the four numbers is 8

In the original condition, there are 4 variables(w,x,y,z), which should match with the number equations. So you need 4 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer. When 1) & 2), standard deviation(d)=root((x-average)^2 is average). So, d=root{[(3-8)^2+(x-8)^2+(y-8)^2+(z-8)^2]/4}, which is always yes and sufficient. Therefore, the answer is C.

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D. _________________

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