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Re: Set A = {22, 27, 11, 12, 2}, and set B: {14, 19, 3, 4, x} [#permalink]
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GreenlightTestPrep wrote:
GreenlightTestPrep wrote:
Set A = {22, 27, 11, 12, 2}, and set B: {14, 19, 3, 4, x}

If x < 0, and the standard deviation of set A = the standard deviation of set B, what is the value of x?

A) -2
B) -3
C) -4
D) -5
E) -6


The test-makers like to test the following concept: If we take a set of numbers with a Standard Deviation of X and add k to each value in the set, the resulting set will still have a Standard Deviation of X.
This should make sense, since Standard Deviation is a measurement of dispersion (i.e., how dispersed the numbers are)

For example, if we take the set {1,2,3} and add 15 to each set member, we get {16, 17, 18}
In both sets, the second number is 1 greater than the first number, and the third number is 1 greater than the second number.
As such, both sets have the same Standard Deviation.

Likewise, if we take the set {7,7,17} and add 11 to each set member, we get {18, 18, 28}
In both sets, the first two numbers are the same, and the third number is 10 greater than the other two numbers.
As such, both sets have the same Standard Deviation.

If we arrange both sets in ascending order we get:
Set A = {2, 11, 12, 22, 27}
Set B = {x, 3, 4, 14, 19}

Notice that each of the 4 biggest numbers in set A are 8 greater than each of the 4 biggest numbers in set B
For example, 27 = 19 + 8
And 27 = 19 + 8
And 22 = 14 + 8
And 12 = 4 + 8
And 11 = 3 + 8

Since the two sets have the same Standard Deviation, it must be the case that 2 = x + 8
Solve for x to get: x = -6

Answer: E

Cheers,
Brent

My approach was different. Since Both sets have same standard deviation, the spreads from medians won't change either, so set 1 => 2-12= -10 therefore set 2 =>x-4= -10 hence x = -6. Is this approach correct as well ?
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Re: Set A = {22, 27, 11, 12, 2}, and set B: {14, 19, 3, 4, x} [#permalink]
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iampratikthorat wrote:

My approach was different. Since Both sets have same standard deviation, the spreads from medians won't change either, so set 1 => 2-12= -10 therefore set 2 =>x-4= -10 hence x = -6. Is this approach correct as well ?


That approach works too :)
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