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If the positive number d is the standard deviation of n, k and p then the standard deviation of n+1, k+1 and p+1 is

(A) d+3 (B) d+1 (C) 6d (D) 3d (E) d

GyanOne is absolutely right about the answer, it is indeed E. I've often heard students say that standard deviation is their least favorite statistic. I, however, think that it should be the opposite. Calculating a square root may seem tedious to some, but the GMAT never asks you to do the actual calculations, so the questions always revolve around the concept. In this case, whether the numbers n, k and p are all increased by 1 or 100, or even 5,000,000, the answer remains the same. The only time the standard deviation is going to increase or decrease is if the numbers are multiplied (or divided, which is really just multiplication by 1/number) by some constant.

A standard deviation question is an opportunity to get easy points because the concept being tested will never be that hard. The important notions are that moving the numbers by a constant changes nothing, multiplying amplifies everything (as you would by pinching an iphone screen) and that sets with the same numbers have standard deviations of 0 (ex: 5,5,5,5).

Quick refresher to calculate standard deviation: 1) Take the average of all numbers. 2) Subtract the average from each element of the set 3) Square this difference (they're all positive now) 4) Add up all these numbers and divide by the number of elements 5) Take the square root of this number. (note: the square is the variance, not tested on the GMAT)

Again, you don't ever have to do this on the GMAT, but the concep can help unlock quick answers to standard deviation questions.
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Re: If the positive number d is the standard deviation of n, k and p then [#permalink]

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19 Feb 2013, 09:42

mun23 wrote:

If the positive number d is the standard deviation of n, k and p then the standard deviation of n+1, k+1 and p+1 is

(A) d+3 (B) d+1 (C) 6d (D) 3d (E) d

Maybe if the theory is not known, one of methods might be to plugin numbers. Assume n,k,p as 1,2,3. Initially, S.D =\(\sqrt{\frac{(1-2)^2+0^2+(3-2)^2}{3}}\) = \(\sqrt{\frac{2}{3}}\). Similarly, S.D for 2,3,4 will be \(\sqrt{\frac{(2-3)^2+0^2+(4-3)^2}{3}}\) = \(\sqrt{\frac{2}{3}}\)

Maybe if the theory is not known, one of methods might be to plugin numbers. Assume n,k,p as 1,2,3. Initially, S.D =\(\sqrt{\frac{(1-2)^2+0^2+(3-2)^2}{3}}\) = \(\sqrt{\frac{2}{3}}\). Similarly, S.D for 2,3,4 will be \(\sqrt{\frac{(2-3)^2+0^2+(4-3)^2}{3}}\) = \(\sqrt{\frac{2}{3}}\)

Thus its the same.

E.

I'm a big fan of plugging in a few numbers if the theory is a little vague. Instead of racking your brain trying to remember some obscure rule (like odd x odd = even/odd?), just try with numbers and you'll see the answer quickly (3 x 5=15, so the product of odd x odd must be odd).

Always a good suggestion to keep in mind that we can proceed this way, thanks Vinaymimani!
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Re: If the positive number d is the standard deviation of n, k and p then [#permalink]

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19 Oct 2014, 00:15

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Re: If the positive number d is the standard deviation of n, k and p then [#permalink]

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25 Oct 2016, 14:29

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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