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List S and List T each contain 5 positive integers, and for

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List S and List T each contain 5 positive integers, and for [#permalink] New post 30 Aug 2006, 12:05
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List S and List T each contain 5 positive integers, and for each list the average (arithmetic mean) of the integers in the list is 40. If the integers 30,40,50 are in both lists, is the standard deviation of the integers in list S greater than the standard deviation of the integers in list T?

(1) The integer 25 is in list S
(2) The integer 45 is in list T
[Reveal] Spoiler: OA

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Last edited by Bunuel on 04 Jul 2013, 02:04, edited 3 times in total.
Added the OA and moved to DS forum.
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 [#permalink] New post 30 Aug 2006, 12:10
I'm going with C here.

1) and 2) both give values for s and t, respectively. To keep the average at 40, we are able to guess the final unknown values in the set.

T = {30, 35, 40, 45, 50}
S = {25, 30, 40, 50, 55}

There is a greater range with S, thereby showing it has a higher standard deviation.
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 [#permalink] New post 30 Aug 2006, 13:42
Looks like C.

St1: S = {25,30,40,50,55}, No information about two values of T.: INSUFF

St2: T = {30,35,40,45,50}, No information about two values of S.: INSUFF

Together:
S = {25,30,40,50,55}
T = {30,35,40,45,50}
SD of T is less because more values are close to mean.: SUFF
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 [#permalink] New post 30 Aug 2006, 14:27
Yep - C.

1) Calculate all members of S, but not of T. Insuff.
2) Calculate all members of T, but not of S. Insuff.

Combine. We have both sets and can get a result.
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 [#permalink] New post 31 Aug 2006, 04:25
Straightforward (C). If we know the (arithmetic) mean, then we can easily solve for the missing number in each list by using the average formula.

Compare averages in each set to determine the SD.
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Re: List S and List T each contain 5 positive integers, and for [#permalink] New post 04 Jul 2013, 01:46
Just curious when the statements are combined

The S.D of Set S will be greater than S.D of set T?

No doubt the answer is C
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Re: List S and List T each contain 5 positive integers, and for [#permalink] New post 04 Jul 2013, 02:07
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fozzzy wrote:
Just curious when the statements are combined

The S.D of Set S will be greater than S.D of set T?

No doubt the answer is C


List S and List T each contain 5 positive integers, and for each list the average (arithmetic mean) of the integers in the list is 40. If the integers 30,40,50 are in both lists, is the standard deviation of the integers in list S greater than the standard deviation of the integers in list T?

The mean of 5 integers is 40 means that the sum of these 5 integers is 5*40=200. The sum of the 3 out of these 5 integers is 30+40+50=120, thus the sum of the remaining 2 integers in each lists is 200-120=80.

(1) The integer 25 is in list S. The 5th integer in S is 80-25=55. We know list S. Not sufficient.
(2) The integer 45 is in list T. The 5th integer in T is 80-45=35. We know list T. Not sufficient.

(1)+(2) We know all terms of each set, thus we can get the standard deviation of each and compare. Sufficient.

Answer: C.

As for your question:

S = {25, 30, 40, 50, 55}
T = {30, 35, 40, 45, 50}

The standard deviation of a set shows how much variation there is from the mean, how widespread a given set is. So, a low standard deviation indicates that the data points tend to be very close to the mean, whereas high standard deviation indicates that the data are spread out over a large range of values.

List T is less widespread thus will have lower standard deviation than that of list S.

Hope it's clear.
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Re: List S and List T each contain 5 positive integers, and for   [#permalink] 04 Jul 2013, 02:07
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