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For the set {6, 15, y}, the standard deviation is x. What is the stand

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For the set {6, 15, y}, the standard deviation is x. What is the stand  [#permalink]

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New post 07 Mar 2017, 02:08
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A
B
C
D
E

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Question Stats:

77% (00:43) correct 23% (01:08) wrong based on 80 sessions

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Re: For the set {6, 15, y}, the standard deviation is x. What is the stand  [#permalink]

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New post 07 Mar 2017, 02:20
1
Bunuel wrote:
For the set {6, 15, y}, the standard deviation is x. What is the standard deviation of the set {10, 19, (y + 4)}?

A. x
B. x + 4/3
C. x + 4
D. x + 12
E. Cannot be determined from the information provided


Note: If we add or subtract a constant to each term in a set: Mean will increase or decrease by the same constant but SD will not change.

constant term added to set {6, 15, y} is 4 so the standard deviation remains x

Hence option A is correct
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Re: For the set {6, 15, y}, the standard deviation is x. What is the stand  [#permalink]

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New post 07 Mar 2017, 09:57
If numbers in sequence increase by the same number then the standard deviation doesn't change. Therefore, the answer is A
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For the set {6, 15, y}, the standard deviation is x. What is the stand  [#permalink]

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New post 08 Mar 2017, 00:39
Bunuel wrote:
For the set {6, 15, y}, the standard deviation is x. What is the standard deviation of the set {10, 19, (y + 4)}?

A. x
B. x + 4/3
C. x + 4
D. x + 12
E. Cannot be determined from the information provided


Official solution from Veritas Prep.

While the abstraction in this problem may make it seem like the new standard deviation cannot be calculated, in fact one core rule about standard deviation should make this one straightforward. Since standard deviation is calculated based on the difference between each term and the mean, if all terms in a set shift by the same amount (here that amount is +4), the standard deviation will remain unchanged. Here, whatever the mean is will simply increase by 4, because each term will have increased by 4. And then each term will be the same distance away from the mean as it was before - the whole scale has simply moved 4 places to the right on the number line. Therefore, because the standard deviation is the same for the new set as it was for the old, the standard deviation is again exactly x.

If you add 4 to each term in the set, the deviation won't change. For instance, take the set {6, 15, 21}

Mean = (6 + 15 + 21)/3 = 14

Distance of each term = 8, 1, 7

Now add 4 to each term, new set {10, 19, 25}, mean = 18

Distance of each term = 8, 1, 7

Voila, same standard deviation

The sets {x, y, z} and {x+k, y+k, z+k} always have the same standard deviation for any value of k.
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Re: For the set {6, 15, y}, the standard deviation is x. What is the stand  [#permalink]

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New post 25 Sep 2018, 23:03
While the abstraction in this problem may make it seem like the new standard deviation cannot be calculated, one core rule about standard deviation makes this one straightforward.

Quote:
Since standard deviation is calculated based on the difference between each term and the mean, if all terms in a set shift by the same amount (here that amount is +4), the standard deviation will remain unchanged.


Here, whatever the mean is will simply increase by 4, because each term will have increased by 4. And then each term will be the same distance away from the mean as it was before - the whole scale has simply moved 4 places to the right on the number line.

Therefore, because the standard deviation is the same for the new set as it was for the old, the standard deviation is again exactly x.

Quote:
Take Away: Since standard deviation is calculated based on the difference between each term and the mean, if all terms in a set shift by the same amount (here that amount is +4), the standard deviation will remain unchanged.

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Re: For the set {6, 15, y}, the standard deviation is x. What is the stand   [#permalink] 25 Sep 2018, 23:03
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