List M (not shown) consists of 8 different integers, each of which is

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.

Thus C

Great explanation, Bunuel! Thank you.

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.

Can you give more question like this to practice

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.

Answer (C)

Check Standard Deviation Questions in our Special Questions Directory.

Also, check our Question Banks for all questions classified by category and difficulty.

Hope it helps.

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).

Responding to a pm:



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

S-1 )Not Sufficient. The key point to understand this question is for Average to remain same the Summation of M = {.......} must be 130-26 = 104. So the possible combination of numbers (22,4) (20,6)...are to be eliminated.
S-2) Not Sufficient
S-T) Once we know in S-1 & S-2 that 22 is not in M, we determine other number as 4. So Yes C is the answer. Wow.

Thanks,
CoolKl

Why can M be something like 12,12,12,12,14,14,14,14.

I chosed E becasue I though the number could repeat.


THanks

List M (not shown) consists of 8 different integers, each of which is in the list shown.

List M (not shown) consists of 8 different integers:
THis doesnot mean that the set consists of 8 integers.
This means that the set will have 8 different integers.
Even (4, 4, 8, 10, 12, 14, 16, 18, 20, )
is a valid set.

SO will the answer then not change?

I say this on the basis of the following questions:

https://gmatclub.com/forum/s-is-a-set-o ... 39444.html

Can any of the experts please clarfiy:

Bunuel
VeritasKarishma
Bunuel
chetan2u
MathRevolution
EgmatQuantExpert[/quote]

You should read a question and a thread more carefully: https://gmatclub.com/forum/list-m-not-s ... l#p1935847

HI Bunuel
What I am actually saying is different from what the other person is saying.

I intend to say that if a set consists of 8 different integers ( 4,6,....22)
then we cannot find out the total number of elements in the set.

For example: (4, 4, 8, 10, 12, 14, 16, 18, 20, )
also consists of 8 different integers but the number of elements is 9.

I hope you see where I am confused.

Thank you

Stmt 1:
Avg of the master list = \((4+6+ 8+ 10+ 12+ 14+ 16 + 18+ 20 + 22)/10 = 13\)
Sum of the elements of the master list = 130

Avg of list M = (Sum of list M) \( / 8 \)
Since, Avg of the master list = Avg of list M
Sum of list M = 13*8=104
Possibilities of elements in list M such that sum=104 i.e. we need to exclude 2 elements whose sum is 26(130-104)
1) { 6, 8, 10, 12, 14, 16, 18, 20} exclude 4, 22
2) { 4, 8, 10, 12, 14, 16, 18, 22} exclude 6, 20 and so on
we get 2 different std deviation values
Not sufficient

Stmt 2:
Exclude 22
Possibilities of elements in list M
1) { 6, 8, 10, 12, 14, 16, 18, 20} exclude 4, 22
2) { 4, 8, 10, 12, 14, 16, 18, 20} exclude 6, 22 and so on
we get 2 different std deviation values
Not sufficient

Combining 1 and 2
Master list = { 4, 6, 8, 10, 12, 14, 16, 18, 20}
sum of master list = 108
sum of list M = 104
We need to exclude 1 element from this master list such that it is 4
only 1 possibility - { 6, 8, 10, 12, 14, 16, 18, 20}

Sufficient
Ans. C

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