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The answer should be (D) i.e. Both I AND II

I.
The SD for a set of number remains same if added or subtracted by a constant. So, (r-s, s-2, and t-2 ) has same SD as (r, s, and t).
This is because the spread between the nos. is not changing. Only the set of nos. is shifting 2 units away.

II
(-s, -r, and -t) are just the mirror image of (s, r and t) on the number line. Thus the set (-s, -r and -t) has the same spread as the set (s, r and t).
Thus the SD for (-s, -r and -t) is same as that of (s, r and t).
Also if a set is added by a constant number its SD remains the same.
So, (-s+s, -r+s and -t+s) has same SD that of (-s, -r and -t)
Thus, (0, s-t, s-r) has same SD as that of (s, r, t)
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Pick easy consecutive numbers and test. Say {s,r,t} is {-2,0,2} so deviation from mean of 0 is 2

I. {-4,-2,0} deviation from mean of -2 is also 2
II. {0, -2-2, -2-0} or {-4,-2,0}, the same as above
III. {|-2|, |0|, |2|} or {0, 2, 2}, both the mean and deviation have changed
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The standard deviation of a set won't change if we add / subtract the same number to / from each element of the set.
If we multiply or divide every term in the set by the same number, the standard deviation will change by the modules of same number. For example, if we multiply each element of the set by 3 or -3, the standard deviation of the set will increase by |3|=3

In I we are subtracting 2 from each term of the set { s, r, t }. SD is the same with that of { s, r, t }
In II we are subtracting s from each term of the set and multiplying by -1. SD is the same with that of { s, r, t }
In III we don't now which elements of the set are negative and which are positive. If all are positive or negative then SD will be the same with that of { s, r, t }. Otherwise it will differ.
So the correct answer is D.
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KarishmaB
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Which of the following has the same standard deviation as {s, r, t}?

I. {r - 2, s - 2, t - 2}
II. {0, s - t, s - r}
III. {|r|, |s|, |t|}

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only


If you add/subtract the same number from each element of a set, the SD does not change.

I. {r - 2, s - 2, t - 2}
Subtracting 2 from each elements will not change the SD.

II. {0, s - t, s - r}

{s - s, s - t, s - r}

Note that whatever the SD of {s, r, t}, the same will be the SD of {-s, -r, -t} because relative distance between them on the number line does not change (all the numbers are just flipped across the Y axis - positive becomes negative, negative becomes positive).
So, when we add s to each term, we still get the same SD.

III. {|r|, |s|, |t|}

This could change the relative distance between them on the number line if r, s and t all do not have the same sign.
e.g.

______________ (-5)_______________________0______________(3)____(4)

becomes

__________________________________________0______________(3)____(4)____ (5)
Much smaller SD now.

Answer (D)

Hello KarishmaB - Typically when we multiply a set by a constant, SD changes. Would -1 be an exception to this?
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VikasBaloni
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Bunuel
Which of the following has the same standard deviation as {s, r, t}?

I. {r - 2, s - 2, t - 2}
II. {0, s - t, s - r}
III. {|r|, |s|, |t|}

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only


If you add/subtract the same number from each element of a set, the SD does not change.

I. {r - 2, s - 2, t - 2}
Subtracting 2 from each elements will not change the SD.

II. {0, s - t, s - r}

{s - s, s - t, s - r}

Note that whatever the SD of {s, r, t}, the same will be the SD of {-s, -r, -t} because relative distance between them on the number line does not change (all the numbers are just flipped across the Y axis - positive becomes negative, negative becomes positive).
So, when we add s to each term, we still get the same SD.

III. {|r|, |s|, |t|}

This could change the relative distance between them on the number line if r, s and t all do not have the same sign.
e.g.

______________ (-5)_______________________0______________(3)____(4)

becomes

__________________________________________0______________(3)____(4)____ (5)
Much smaller SD now.

Answer (D)

Hello KarishmaB - Typically when we multiply a set by a constant, SD changes. Would -1 be an exception to this?

  • If we multiply (or divide) all the numbers in a set by a non-zero constant, the standard deviation will be respectively multiplied or divided by the absolute value of that constant. For example, if the SD of \(\{a, \ b, \ c\}\) is \(s\), then the SD of \(\{2a, \ 2b, \ 2c\}\) will be \(2s\) and the SD of \(\{-3a, \ -3b, \ -3c\}\) will be \(s*|-3|=3s\).
  • If we add (or subtract) a constant to all the numbers in a set, the standard deviation will NOT change. For example, if the SD of \(\{a, \ b, \ c\}\) is \(s\), then the SD of \(\{a+4, \ b+4, \ c+4\}\) or of \(\{a-1, \ b-1, \ c-1\}\) will also be \(s\).

Therefore, if we multiply (or divide) all the numbers in a set by -1, the standard deviation will remain unchanged. This is because the standard deviation is multiplied by the absolute value of the factor, which is |-1| = 1 in this case.

Hope this helps.
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