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Re: Data sufficiency OG 13 - Statistics [#permalink]
15 Oct 2012, 19:22

1

This post received KUDOS

pnf619 wrote:

4,6,8,10,12,14,16,18,20,22

List M (not shown) consists of 8 different integers, each of which is in the list shown. what is the standard deviation of the numbers in the list M?

1) The average ( arithmetic mean) of the numbers in the list M is equal to the average of the numbers in the list shown. 2) list M is does not contain 22.

Stuck on this question and taking a lot of time to solve it any good solutions for this question?

Let us name this list as N, which contains 4,6,8,10,12,14,16,18,20,22. AMean is = 13 Now we need to prepare a list M. statememnt 1--> avg of M = avg of L M could be anything but has an avg of 13...so ---> not sufficient stmt 2 ---> M does not has 22 . --->not sufficient

using both 1 and 2 --> avg=13 and no 22 in the list. we have only one list left: 6,8,10,12,14,16,18,20 which has avg of 13.

Re: Data sufficiency OG 13 - Statistics [#permalink]
16 Oct 2012, 02:07

1

This post received KUDOS

The question states that List M has 8 numbers to be picked from 10. 12+14/2 = 13 since 10 numbers the sum would be 130. since statement 1 says that we need to have an equal average of 13. now the sum needs to be 13*8=104.

so we have to reduce the sum by 26 which can be done in many different ways but just pointing out 2 cases 6,8,10,12,14,16,18,20 4,8,10,12,14,16,18,22

both sets have different SD so insufficient

statement 2 says that you can't pick 22 4,6,8,10,12,14,16,18 4,6,8,10,12,14,16,20

again insufficient

now when we combine 1 and 2 we can't pick 22 so our list is 4,6,8,10,12,14,16,18,20

but in our list M we need the sum to add up to 104 this can only be done by eliminating 4 6,8,10,12,14,16,18,20

so the only list available is the one mentioned above so the answer is C

Re: List M (not shown) consists of 8 different integers, each [#permalink]
16 Oct 2012, 04:35

13

This post received KUDOS

Expert's post

4, 6, 8, 10, 12, 14, 16, 18, 20, 22

List M (not shown) consists of 8 different integers, each of which is in the list shown. What is the standard deviation of the numbers in list M ?

Given list consists of 10 evenly spaced integers. Mean=(First+Last)/2=13 and Sum=(Mean)*(# of terms)=130.

Given that list M is obtained by removing 2 integers from the list shown.

To determine the standard deviation of list M we must know which 2 integers were removed.

(1) The average (arithmetic mean) of the numbers in list M is equal to the average of the numbers in the list shown --> the mean of list M is also 13. Thus the sum of the integers in list M is 13*8=104, which means that the sum of the 2 integers removed is 130-104=26. The 2 integers removed could be: (4, 22), (6, 20), ..., (12, 14). Not sufficient.

(2) List M does not contain 22. We know only one of the numbers removed. Not sufficient.

(1)+(2) From (1) we know that the the sum of the 2 integers removed is 26 and from (2) we know that one of the integers removed is 22. Therefore the second integer removed is 26-22=4. List M consists of the following 8 integers: {6, 8, 10, 12, 14, 16, 18, 20}. So, we can determine its standard deviation. Sufficient.

Re: List M (not shown) consists of 8 different integers, each [#permalink]
04 May 2014, 13:47

Bunuel wrote:

Bumping for review and further discussion.

Hi Bunuel,

I have a general question on Standard Deviation. I realize that SD is the spread from the mean, but what I have a hard time understanding is if "weight averages" come into play. Let's assume the list is [10,10,14,18,18] -- the spread is from 10 to 18 so the deviation is 4 to the right and 4 to the left. Correct?

Now if we assume that the list is [10,13,14,15,25] -- without calculating(since the gmat won't ask us to calculate SD if i'm not mistaken, which one has the higher SD? The second list obviously has a wider range(10 to 25) but the numbers are bunched up closer. I guess, what i'm asking is, what carries more weight? Have a wider range or have multiple numbers on the edges(albeit a smaller range).

Re: List M (not shown) consists of 8 different integers, each [#permalink]
05 May 2014, 00:08

Expert's post

russ9 wrote:

Bunuel wrote:

Bumping for review and further discussion.

Hi Bunuel,

I have a general question on Standard Deviation. I realize that SD is the spread from the mean, but what I have a hard time understanding is if "weight averages" come into play. Let's assume the list is [10,10,14,18,18] -- the spread is from 10 to 18 so the deviation is 4 to the right and 4 to the left. Correct?

Now if we assume that the list is [10,13,14,15,25] -- without calculating(since the gmat won't ask us to calculate SD if i'm not mistaken, which one has the higher SD? The second list obviously has a wider range(10 to 25) but the numbers are bunched up closer. I guess, what i'm asking is, what carries more weight? Have a wider range or have multiple numbers on the edges(albeit a smaller range).

Hope my question makes sense.

Thanks

Neither alone. For example, {1, 8} has larger stander deviation than {1, 3, 5, 7, 9} (notice that the first set has smaller range and less terms then the second one).

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.

For, {10, 10, 14, 18, 18}: mean=14, and the deviations from the mean are 4, 4, 0, 4, 4. For, {10, 13, 14, 15, 25}: mean=15.4, and the deviations from the mean are 5.4, 2.4, 1.4, 9.6, 49.6.

Since the second set is a bit more widespread then the first one, then it must have larger standard deviation.