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

Thanks for the explanation Ian. C is the right answer.

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.

so sd = (5 + 5)/2 = 2.5 > 2

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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