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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]

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30 Aug 2006, 13:05
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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, 03:04, edited 3 times in total.
Added the OA and moved to DS forum.

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30 Aug 2006, 13:10
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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30 Aug 2006, 14:42
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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30 Aug 2006, 15:27
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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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31 Aug 2006, 05:25
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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]

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04 Jul 2013, 02:46
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]

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04 Jul 2013, 03:07
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Expert's post
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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Re: List S and List T each contain 5 positive integers, and for [#permalink]

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03 Jan 2015, 10:46
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Re: List S and List T each contain 5 positive integers, and for [#permalink]

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25 Jan 2016, 00:10
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List S and List T each contain 5 positive integers, and for [#permalink]

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15 Jul 2016, 02:37
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 [#permalink]

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15 Jul 2016, 04:06
C

1)value of S is given not T.insuff
2)value of T is given not S.insuff

1+2

avg=40 for both

but spread of mean from each value is greater in S than T

so S>T

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

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11 Feb 2017, 10:08
Bunuel wrote:
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.

Calculating SD with formula Range/4, Is this correct?
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Re: List S and List T each contain 5 positive integers, and for [#permalink]

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11 Feb 2017, 11:55
anairamitch1804 wrote:
Bunuel wrote:
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.

Calculating SD with formula Range/4, Is this correct?

No. Where is the SD calculated as range/4?
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Re: List S and List T each contain 5 positive integers, and for   [#permalink] 11 Feb 2017, 11:55
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