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Re: Data sufficiency OG 13 - Statistics [#permalink]

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15 Oct 2012, 20:22

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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]

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16 Oct 2012, 03:07

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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

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]

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04 May 2014, 14:47

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).

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.
_________________

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 ?

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

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

It's a good question and can be easily solved using your understanding of mean of AP.

The list shown has 10 equally spaced numbers. Their mean will be the average of middle two numbers i.e. average of 12 and 14 which is 13. List M has 8 of these 10 numbers. We need the SD of list M.

(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 13. But we don't know how the numbers of list M deviate from the mean. We can select 8 numbers in different ways to get mean of 13.

(2) List M does not contain 22. We are left with 9 numbers from which we select 8. The SD will be different depending on the numbers we select.

Using both, we have 9 numbers whose mean must be 13. One easy way we know in which we can select the numbers is drop 4 to get 8 equally spaced numbers whose mean will be 13. List M - (6, 8, 10, 12, 14, 16, 18, 20) Can you get the same mean by dropping some other number and keeping 4? Think about it - it is not possible. The number of numbers must stay 8. If you replace any other number by 4, the total sum will change which will change the mean. Hence, the only way to select list M is this one. We can easily find the SD here so both statements together are sufficient.

Re: List M (not shown) consists of 8 different integers, each [#permalink]

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16 May 2015, 09:18

The list consists of 10 numbers, which means we're taking away two of them.

(1) Tells us that the two numbers that have been removed were equally far from the mean, but opposite directions. e.g. 4 and 22, 6 and 20, or 8 and 18 etc. It's not sufficient to answer the question as the removal of higher numbers would yield a smaller spread and thus a smaller SD, than would more centric numbers.

(2) Tells us that 22 was one of the numbers erased. This does not help us as we would need to know what the other number is.

(1)+(2) In addition to the reasoning above, we can conclude that the second number erased was 4. This is based on the restrictions in regards to the possible and distinct pairs from (1) along with one of the members of the pair from (2).
_________________

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Re: List M (not shown) consists of 8 different integers, each [#permalink]

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19 May 2016, 23:55

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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 ?

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

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

Responding to a pm:

Quote:

I solved this question using mean and median. When the numbers in a set are equally spaced, mean is equal to the median. Therefore in set {4,6,8,10,12,14,16,18,20,22}, (12+14)/2 = 13 is the median and the mean. Therefore I assumed that 12 and 14 would remain in the set and as they form the median they would remain the middle values.

Is it correct to assume this question in the manner I did?

Mean is equal to median in the list shown. What says that mean will be equal to median in list M too? Also, if mean = median, it doesn't mean that the set MUST be equally spaced. Even if 12 and 14 are not there in list M, the median of the rest of the set will still be 13.

M = {4,6,8,10,16,18,20,22} Median = (10+16)/2 = 13
_________________

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