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Director  Joined: 10 Feb 2006
Posts: 549
List S and list T each contain 5 positive integers, and for each list  [#permalink]

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Difficulty:   15% (low)

Question Stats: 83% (01:49) correct 17% (01:20) wrong based on 305 sessions

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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, and 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.

Originally posted by alimad on 30 Aug 2006, 12:05.
Last edited by Bunuel on 20 Oct 2019, 00:18, edited 4 times in total.
Renamed the topic and edited the question.
Manager  Joined: 27 Jul 2006
Posts: 158
Re: List S and list T each contain 5 positive integers, and for each list  [#permalink]

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1
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.
Senior Manager  Joined: 06 May 2006
Posts: 489
Re: List S and list T each contain 5 positive integers, and for each list  [#permalink]

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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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CEO  B
Joined: 29 Jan 2005
Posts: 3519
Re: List S and list T each contain 5 positive integers, and for each list  [#permalink]

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1
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.
Director  Joined: 29 Nov 2012
Posts: 677
Re: List S and list T each contain 5 positive integers, and for each list  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 61358
Re: List S and list T each contain 5 positive integers, and for each list  [#permalink]

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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.

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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Director  B
Joined: 04 Jun 2016
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Re: List S and list T each contain 5 positive integers, and for each list  [#permalink]

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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

Using stimulus we know
S={30,40,50,-,-,} two unknown ; Mean is 40
T={30,40,50,-,-} two unknown ; Mean is 40

Lets use the quick "EYEBALL SD FROM MEAN" method .
Now the set which will have values far from 40 will have a greater SD

Stament(1) The integer 25 is in list S
It means our set is now complete S={25,30,40,50,55}
But it tells us nothing about element of other set.
INSUFFICIENT

(2) The integer 45 is in list T
Meaning now out set T is complete t={30,35,40,45,50}
But it tells that nothing about element of set S
INSUFFICIENT

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

Lets use the quick "EYEBALL SD FROM MEAN" method .
SD of S will be greater since it extreme values are more spread out from the mean (40-25 =15 and 40-55=-15) Exact SD will be 12.7
SD of T will be lower since it extreme values are less spread out from the mean (40-30 =10 and 40-50=-10) Exact SD will be 7.9

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Re: List S and list T each contain 5 positive integers, and for each list  [#permalink]

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Re: List S and list T each contain 5 positive integers, and for each list  [#permalink]

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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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_________________ Re: List S and list T each contain 5 positive integers, and for each list   [#permalink] 20 Oct 2019, 00:16
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