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List S and List T each contain 5 positive integers, and for [#permalink]
30 Aug 2006, 12:05

00:00

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C

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Difficulty:

25% (medium)

Question Stats:

73% (01:54) correct
27% (00:48) wrong based on 51 sessions

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

Re: List S and List T each contain 5 positive integers, and for [#permalink]
04 Jul 2013, 02:07

1

This post received KUDOS

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.

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.