tough statistics question

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tough statistics question

by gmatquant25 » Thu Apr 18, 2013 9:27 am
If 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 largest negative element of set S is -1

please elaborate the answer ..

OA C

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by [email protected] » Thu Apr 18, 2013 9:48 am
gmatquant25 wrote:If 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 largest negative element of set S is -1
Statement 1: Exactly half of all the elements of set S are positive.
Hence, the rest of all the elements are either all negative or some negative and some zeroes or all zeroes. Consider the following cases,
  • {-2, -2, 1, 2} --> Median = (-2 + 1)/2 = -0.5 --> Negative
    {-2, -1, 1, 2} --> Median = (-1 + 1)/2 = 0 --> Not negative
Not sufficient

Statement 2: This means -1 is a element of the S. But from this we cannot determine whether the median is negative or not.

Not sufficient

1 & 2 Together: As we know that exactly half of the elements of S are positive and S has even number of elements, the median of S will be (smallest positive element + the integer just smaller than the smallest positive element)/2.

Now, smallest positive integer that can be in S is 1.
And, the integer just smaller than the smallest positive element of S is either 0 or -1.
Hence, the smallest value of the median is (1 - 1)/2 = 0
So, the median is always non-negative.

Sufficient

The correct answer is C.
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by BenMiller » Thu Apr 18, 2013 9:53 am
First, we know we have an even number of integers, so the median wil be the AVERAGE of the two middle terms.

Looking at 1) we know that half are positive, so half must be negative or 0. To know if the median is negative, we need to know which integer that goes into the median "weighs more", or is further from 0. Whichever term is furthest from 0 will "pull" the sign of the median in that direction.

Since we don't know anything about either of the two middle terms, 1) is therefore not sufficient.

Examining 2) we are provided a single element of the set with no context, so we can't know if -1 is in the middle, or on one of the ends of the set. Again, insufficient.

Putting them together, we know that the median is the average of -1 (the highest negative) and some x (the lowest positive). Referring back, we still need the distance from 0 of the positive term in order to figure out the sign. So our answer is E.

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by [email protected] » Thu Apr 18, 2013 6:12 pm
BenMiller wrote:Putting them together, we know that the median is the average of -1 (the highest negative) and some x (the lowest positive). Referring back, we still need the distance from 0 of the positive term in order to figure out the sign. So our answer is E.
Hi Ben,

First of all, the median need not be average of -1 and the lowest positive integer. It can very well be average of 0 and the lowest positive integer. But for the minimum possible median, it will average of -1 and the lowest positive integer.

And we do not need to know the lowest positive.
Even if we make the lowest positive element as small as possible, it'll be 1 (as all the elements are integers), and the smallest median in that case will be (-1 + 1)/2 = 0, which is a non-negative.
Hence, we can conclude that median will be always non-negative.

However, your solution would have been perfect if it was not mentioned that all the elements are integers.
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by mariofelixpasku » Sun Apr 21, 2013 12:23 am
ok got it

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by vipulgoyal » Sun Apr 21, 2013 9:23 pm
Experts please suggest

If 0 is positive int ( As shoown in ans (-1 + 1)/2 = 0 +ve) then median could be -ve
of series -2,-1,0,1
-1 + 0 / 2 = -1/2

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by [email protected] » Sun Apr 21, 2013 9:30 pm
vipulgoyal wrote:If 0 is positive int ( As shoown in ans (-1 + 1)/2 = 0 +ve) then median could be -ve
of series -2,-1,0,1
-1 + 0 / 2 = -1/2
0 is neither positive nor negative.
I've written : '{-2, -1, 1, 2} --> Median = (-1 + 1)/2 = 0 --> Not negative'
That doesn't mean it is positive.

Hope that helps.
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