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# S and T are sets of numbers. The standard deviation of the

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S and T are sets of numbers. The standard deviation of the [#permalink]

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22 Jul 2006, 05:31
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S and T are sets of numbers. The standard deviation of the elements of set S is q. Is the standard deviation of S U T greater than q?

(1) The range of S U T is different from the range of S.
(2) There is only one element in T, and it is twice the arithmetic mean of the elements in S.

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22 Jul 2006, 09:03
Note: U means union

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22 Jul 2006, 09:40
Quote:
The range of S U T is different from the range of S.

If range of union is different then standard deviation will also defer.

My answer is A

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22 Jul 2006, 13:47
capri wrote:
Quote:
The range of S U T is different from the range of S.

If range of union is different then standard deviation will also defer.

My answer is A

It certainly differs, but read the question carefully

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Re: DS: Two sets [#permalink]

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22 Jul 2006, 14:02
kevincan wrote:
S and T are sets of numbers. The standard deviation of the elements of set S is q. Is the standard deviation of S U T greater than q?

(1) The range of S U T is different from the range of S.
(2) There is only one element in T, and it is twice the arithmetic mean of the elements in S.

1. range differs means it can only increase therefore SD increases

suff

2. one element is 2times more than mean - therefore sd increases ---SD will not increase if the lone element in T is equal to the mean

So D

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Re: DS: Two sets [#permalink]

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22 Jul 2006, 15:05
kevincan wrote:
S and T are sets of numbers. The standard deviation of the elements of set S is q. Is the standard deviation of S U T greater than q?

(1) The range of S U T is different from the range of S.
(2) There is only one element in T, and it is twice the arithmetic mean of the elements in S.

going with C.

Just starting out with math, so am just trying to nail down the concepts.

1: The range can only change in a set if the difference between the greatest elements in the set changes. Since the new set SUT contains all the elements of the old set S, the range of the new set, if different, must be greater, since it cannot be smaller. However we are concerned not with range but with SD. There may be 5 elements in S and 1000 elements in T. One element in T is greater than the largest elememt in S or lesser than the smallest in S, increasing the range. This one element will certainly cause the SD to increase. The other 999 elements may actually be the same value as the arithmetic mean, thereby reducing the SD.

We cannot say from 1, if the SD will increase or decrease, though we can say that the range will increase.

2: There is a single element in T which is twice the average. Consider a set as follows:-

{ 0, 10000}

Now add another element which is twice the mean, i.e. 10000
Since the question does not say the sets are disjoint, this is possible.
Set SUT = S = { 0, 10000 }, and the SD does not change.

Consider another set S {0,10,20} mean = 10, SD = 5.something.
Add an element twice the mean i.e. 20. SD increases.
Therefore 2 is incufficient.

Consider 1 and 2 together. If there is one element added, and this causes the range to increase, it cannot be an element already in S. Then, SD will increase.

Will go with C

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22 Jul 2006, 15:06
C

St1: If range increases then SD may increase or decrease depending on how the mean is affected. Data points may lie farther than mean or closer to mean. INSUFF

St2: That one point may increase or decrease the range and can also increase or decrease the mean.: INSUFF
Case1: S = {-1,0,0,0,1} Mean = 0 So T = {0}
Variance of S = 2/4
Variance of T = 0
Variance of (S and T) = 2/5

Case2: S = {2,2,2,2} means = 2 so T = {4}
Variance of S = 0
Variance of T = 0
Variance of (S and T) is greater than 0 because mean is 12/5 and more points are far scattered to mean.

Combined:

It means one point of T will increase the rangle of (S and T) and will increase the SD.
Case1 of statement is out and case 2 is in. that means SD will increase.

NOTE: Variance = SD^2
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Re: DS: Two sets [#permalink]

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23 Jul 2006, 11:43
kevincan wrote:
S and T are sets of numbers. The standard deviation of the elements of set S is q. Is the standard deviation of S U T greater than q?

(1) The range of S U T is different from the range of S.
(2) There is only one element in T, and it is twice the arithmetic mean of the elements in S.

its B for me.

1. difference in range could increase or decrease the SD. nsf.
2. if T has only one element, then S U T has either 1 or 0 element (since we know from the question that set S has more than 1 element). in both cases the SD is 0. so suff.

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Re: DS: Two sets [#permalink]

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23 Jul 2006, 22:57
kevincan wrote:
S and T are sets of numbers. The standard deviation of the elements of set S is q. Is the standard deviation of S U T greater than q?

(1) The range of S U T is different from the range of S.
(2) There is only one element in T, and it is twice the arithmetic mean of the elements in S.

Will go with B.. B tells that we adding another element that's twice the mean.
This means we are adding square of mean to variance.. This means standard deviation will increase..

Just B..

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24 Jul 2006, 00:52
Why are (1) and (2) sufficient together?

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24 Jul 2006, 07:43
If the range of SUT is same with S then it means we are adding one point (T) within the range of S. The points are less scattered. In other words SD would become smaller.

If the range of SUT is different from S then it means we are adding one point (T) outside the range of S. The points are more scattered. In other words SD would be greater.
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24 Jul 2006, 09:56
E.

A is not sufficient.
I only present an example, in which q decreases: T=(1, 1, 1, 1, 1, 1, 1,1 ....,1) 1*10^12 times; S=(4, 8, 12). Then the standard deviation surely decreases.

B. is not sufficient too.
For example:
S=(0,0,0) T=0. Then q does not increase.
(examples, where q increases are easy to find)

Both statements together are not sufficient.

[Note: It is sufficient to find an example and a counterexample; this sketch is not necessary)

To provide a general insight is somewhat complicated (I can't use symbols here).

Sketch:
We are able to compute the new average (na).
Consider the formula of the new standard deviation:
1/(n+1)*sum(x(i) - x(bar))^2 = 1/(n+1)*sum((x(j)-x(bar))^2)+1/(n+1)*x(bar)^2 = ...
We can compare this term with the old term.
The new standard deviation could be smaller or bigger than the old st. d (just plug-in some numbers).

OE (Short-Cut)?

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24 Jul 2006, 11:38
Yours will do, though with a simpler counterexample:

Any set made up of many slightly postive numbers and one large negative number, a set that has a mean that is slightly positive, too.

Ex The 9-element set (-3,1.5,1.5,1.5,...,1.5) mean 1
variance (4^2+ 8(0.5)^2)/9= 2

If we include the element 2 (twice the mean of S), new mean is 1.1

variance= ((4.1)^2+8(0.4)^2+ (0.9)^2)/10 which is less than (17+1.3+1)/10, which is less than 2.

Admittedly, this is not a fair question, but a lot of people fell for A, which is disconcerting! OA=E

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24 Jul 2006, 11:38
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# S and T are sets of numbers. The standard deviation of the

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