What is the standard deviation of a set of numbers

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What is the standard deviation of a set of numbers whose mean is 20?

(1) The absolute value of the difference of each number in the set from the mean is equal

(2) The sum of the squares of the differences from the mean is greater than 100

OA:E

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by ceilidh.erickson » Sat Oct 01, 2016 6:32 pm
This is not a GMAT-like question at all.

The GMAT expects you to have a general understanding of what standard deviation means: it's a way of expressing how far spread out, on average, the data is from the mean. The GMAT does not expect you to know how that is calculated. The most that the GMAT expects you to know is that standard deviation is related to "variance" (see #D31 in OG2016 or 2015). You will not need to know how that relates to absolute value of difference, or sum of squares, etc.

But in any event, here's how to approach it:

What is the standard deviation of a set of numbers whose mean is 20?

(1) The absolute value of the difference of each number in the set from the mean is equal


All this is telling us is that each term is equally far from 20. Consider two scenarios:

a) Our set consists of all 20's: [20, 20, 20, 20]. The standard deviation would be 0.

b) Our set consists of mirrored pairs of numbers equally far from 20: [19, 19, 21, 21] or [-10, -10, 50, 50]. These two different sets would give us very different standard deviations.

Insufficient.

(2) The sum of the squares of the differences from the mean is greater than 100

This is the definition of variance. The standard deviation would be the square root of this number. So statement 2 tells us that the SD is less than 10, but that doesn't help us. Insufficient.

If we combine statements 1 and 2, we still have no further information on what the SD is. Insufficient.

The answer is E.
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by Jay@ManhattanReview » Tue Dec 20, 2016 5:12 am
NandishSS wrote:What is the standard deviation of a set of numbers whose mean is 20?

(1) The absolute value of the difference of each number in the set from the mean is equal

(2) The sum of the squares of the differences from the mean is greater than 100

OA:E

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We are asked to deduce SD, given that mean = 20.

S1: Case 1: Each element of the set = 20, so the set is: {... 20, 20, 20, ...}. The absolute value of the difference of each number from the mean = 0. Despite not knowing the number of terms in the set, we can conclude that SD = 0.

Case 2: No element of the set = 20, but its mean = 20. Say there are 2n number of elements in the set: n number of elements = (20 - x) and another n number of elements = (20 + x). Thus, the absolute value of the difference of each number from the mean = x. Since we do not know the value of x, S1 is insufficient.

S2: The falls under Case 2 of S1. All it states that 2n*x^2 > 100.

We know that standard deviation is the mean of squared deviations.

SD = [(Sum of the squares of the differences)/ number of terms]^(1/2) = [(2n*x^2)/2n]^(1/2) = [x^2]^(1/2) = x. Since we do not know the value of x, S2 is insufficient.

S1 and S2: Since Case 2 is common for S1 and S2, we cannot get anything even after combining S1 and S2. Insufficient.

OA: E

-Jay

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by GMATGuruNY » Tue Dec 20, 2016 6:10 am
NandishSS wrote:What is the standard deviation of a set of numbers whose mean is 20?

(1) The absolute value of the difference of each number in the set from the mean is equal

(2) The sum of the squares of the differences from the mean is greater than 100
Statements combined:

Case 1: 10, 30
Mean = (10+30)/2 = 20.
Absolute difference from the mean = |20-10| = |20-30| = 10.
Sum of the squares of the differences from the mean = (20-10)² + (20-30)² = 200.

Case 2: 0, 40
Mean = (0+40)/2 = 20.
Absolute difference from the mean = |20-0| = |20-40| = 20.
Sum of the squares of the differences from the mean = (20-0)² + (20-40)² = 800.

Since the data points in Case 2 deviate more from the mean of 20 than do the data points in Case 1, the two sets have different standard deviations.
Since the standard deviation can be different values, INSUFFICIENT.

The correct answer is E.
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