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mun23
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 \(\sqrt{\frac{(2-3)^2+0^2+(4-3)^2}{3}}\) = \(\sqrt{\frac{2}{3}}\)

Thus its the same.

E.
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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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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

Standard deviation depends on the relative placement of numbers with respect to each other only.

So standard deviation in these two cases is the same :

--------n---k-----p-----------
----------n---k-----p---------

Hence the standard deviation of the new set will be d only.

Answer (E)
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Though two great instructors from Veritas Prep have already commented on this, I will take the liberty to Quote the "Map Strategy" from Veritas Prep Book to help remember how SD behaves when a constant is added/subtracted/Multiplied/divided to a set. I hope you find it helpful.

Map Strategy for Standard Deviation (Courtesy - Veritas Prep Statistics and combinatorics book)

  • Adding or subtracting a constant from each element in the set has no effect on the standard deviation. Shifting a map does not change the distances.
  • Multiplying each term of a set by a number with an absolute value greater than 1 increases the standard deviation, while multiplying by a number with an absolute value less than 1 decreases the standard deviation. Increasing the scale of the map increases distances, while reducing the scale shrinks them.
  • Dividing each element in a set by a number with an absolute value greater than 1 decreases the standard deviation, while dividing by a number with an absolute value less than 1 increases the standard deviation. Increasing the scale of the map increases distances, while reducing the scale shrinks them.
  • Changing the sign of all elements in the set or multiplying by -1 has no effect on the standard deviation.

Applying the "Map Strategy" to this question we see that a constant is added to every element of the set. Hence, the SD will remain unchanged. The Required answer is E.
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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

Standard deviation of a set will not change if we add the same number to every member of the set. So adding 1 to each of n, k, and p will not change standard deviation d. E)
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