DS- Statistics - reopening topic

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DS- Statistics - reopening topic

by venmic » Sun Jul 10, 2011 3:30 pm
Set S consists of even number of integers. Is the median of set S negative?

1. Exactly half of all elements of set S are positive
2. The biggest negative element of set S is -1

Reopeningnew thread ... most of the threads are confusing and have not given the right answer

I think the right answer is E

please explain

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by navami » Sun Jul 10, 2011 7:44 pm
Ans should be C.
Both together possible
Lets say set looks like the below ...

[-1 2 ] => Median +ve
or
[ -4 -1 3 9 ] = > median +ve
or
[ -1 0 1 4 ] => median +Ve
so if we consider option 1 and 2 together we know the median can never go negative
This time no looking back!!!
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by 6983manish » Sun Jul 10, 2011 7:49 pm
I will go with C.

Solution ::

Option 1. set of even numbers integers & exactly half are positive.
-4 , -3 , 2 , 1 or -4 , -3 , 5 , 7 .. both satisfy the condition but do not give us definitive answer.

Hence , A and E are out.

Option 2. set of even numbers integers & largest negative integer is -1.

-4, -3 , -1 , 2 , 4, 7 / -4, -3 , -2, 1 / -4 , -3 , -1 , 0 , 1, 2 / -1, 1 , 2,4

All above satisfy the condition but are not giving us the answers.

Hence B is out.

Combining 1 & 2.
set of even numbers integers & largest negative integer is -1 & exactly half are positive.

-4 , -3 , -1 , 4, 5 , 6 / -4, -3 , -1 , 1 , 4 , 5

Both satisfy the condition and give us the answer that it cannot be negative.

Hence C.

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by srikanthb.69 » Tue Jul 12, 2011 6:52 am
what about

-4 , -3 , -1 , 4, 5 , 6 / -4, -3 , -1 ,1 , 4 , 5 --give median +ive


-4, -3 , -1 ,0 , 4 , 5 -- give median -ive so I think the answer is E.

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by Frankenstein » Tue Jul 12, 2011 8:10 am
srikanthb.69 wrote:what about

-4 , -3 , -1 , 4, 5 , 6 / -4, -3 , -1 ,1 , 4 , 5 --give median +ive


-4, -3 , -1 ,0 , 4 , 5 -- give median -ive so I think the answer is E.
Hi,
There are only 2 positive elements(4,5) in your example instead of 3. So, that example is invalid.

If there are 2n elements and the elements are placed in increasing order,, median is average of nth element and (n+1)th term.
As everyone here agrees that using either statement is not sufficient,
Combining (1) and (2):
nth term can be either -1 or 0
All the terms from (n+1)th term are positive integers. So, (n+1)th term can be any positive integer
So, sum of nth term and (n+1)th term is at least zero when nth term is -1 and (n+1)th term is 1.
So, median is at least 0.
Hence, it can never be negative

Hence, C
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by Geva@EconomistGMAT » Tue Jul 12, 2011 11:10 pm
Frankenstein wrote:
srikanthb.69 wrote:what about

-4 , -3 , -1 , 4, 5 , 6 / -4, -3 , -1 ,1 , 4 , 5 --give median +ive


-4, -3 , -1 ,0 , 4 , 5 -- give median -ive so I think the answer is E.
Hi,
There are only 2 positive elements(4,5) in your example instead of 3. So, that example is invalid.

If there are 2n elements and the elements are placed in increasing order,, median is average of nth element and (n+1)th term.
As everyone here agrees that using either statement is not sufficient,
Combining (1) and (2):
nth term can be either -1 or 0
All the terms from (n+1)th term are positive integers. So, (n+1)th term can be any positive integer
So, sum of nth term and (n+1)th term is at least zero when nth term is -1 and (n+1)th term is 1.
So, median is at least 0.
Hence, it can never be negative

Hence, C
EDIT: Agree here. Stat. (1) divides the set into two halves: half positive, half non-positive (zero or smaller). stat. (2) presents the greatest possible negative term as -1, but still allows the greatest term in the non-positive half to be zero. Thus, there are two scenarios:

negatives...-1, positives (where -1 is also the greatest of its group)

OR

negatives...-1, 0, positives (where the non-positive group includes -1 as the greatest negative AND zero as the greatest of the group)

for scenario 1, the smallest possible median will be zero: the smallest positive integer is 1, and the average of -1 and 1 is zero. Any positive integer greater than 1 will result in a greater median, so the answer to the question "is the median negative?" is a definite "no", which is sufficient.

For scenario 2, the smallest possible median will be positive, since it is the average of zero and a positive integer. Again, the answer to the question stem is a definite "no", which is sufficient.

thus, the combination of statements is sufficient to determine that the answer to the question stem is "no". The answer is C.
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by AbhiJ » Thu Aug 25, 2011 2:20 am
Lets assume that we have 10 elements in the set.

From 1 -> 5 elements are positive, let them be 1,2,3,4,5.

From 2 -> Lets assume we have 4 elements all equal to -1.

Take one element to be equal to 0.

Thus the median of the set will be (5th element + 6 element)/2.0

= (0+1)/2 = 0.5

Hence the answer is E.

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by tuanquang269 » Thu Aug 25, 2011 2:46 am
In this case, all of you have to determine clearly MEDIAN and MEAN (average number). I saw that someone consider between C & E answer.

Here is my answer, first, median is the middle value. If a set have even number of element, it will equal to the average of 2 middle value

Let's take some example:

-5, -4, -1, 0, 1, 2, 3, 4 => the median is average of 0 & 1 equal to 0.5, right

-4, - 3, -1, 1, 2, 3 => the median is 0 (average of -1 and 1), right

=> the median is NOT negative number

Moreover, I didn't find the other cases that out of 2 cases above. If you find another, just read carefully question again. WARNING with yes/no question: Is the median of set S negative (or NOT)?

So, the answer is C

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by navami » Thu Aug 25, 2011 2:49 am
Thanks Geva
This time no looking back!!!
Navami